Towards a Renaissance
of
Curved Spanning Structures
by
HUNTER
THOMAS
B.A.,
Princeton
1978
HARTSHORNE
University
in partial fulfil lment
requirements for the
degree of
Master of Architecture
at the
Massachu setts Institute of Technology
June, 1983
Submit ted
of the
@
Hunter
Thomas
1983
Hartshorne
The author hereby grants to M.I.T. permission to
reproduce and to distribute copies of this document
in whole or in part.
Signature
author , . .
of
Certified by . . . v,
Waclaw P. Zalewski,
Thesis Supervisor
Accepted by
Jan Wampler,
Departmental
.
. -- 0
May
Architecture,
Department of
.
a
.
.
.
,-r-if-ex.
6, 1983
-- -. . .
Professor of
.%
. .
.
- .e
-
. -
- -
Chairman
ommittee
for
Graduate
MASSACHUSETS INSTITUTE
OF TECHNOLOGY
MAY 12 61983
LIBRARIES
. -
. ..
.
.
.
Structures
Students
-
- -
-
-
-
II
Towards a Renaissance
of
Curved Spanning Structures
by Thomas Hunter Hartshorne
Submitted to the Department of Architecture on May 6, 1983 in
partial fulfillment of the requirements for the degree of
Master of Architecture.
Abstract
Today, as always, there is a great need for expressiveness in
design. There are a multitude of forms which, over the years, have
been stripped from the designer's repertoire in the name of economics.
There exists a sub-group amongst the lost forms which can be
categorized under the heading, "Curves." Where have all the curves
gone?
The Dutch de Stijl and the German Bauhaus movements did much
to make things unpleasant for the curve in the twentieth century.
The ever increasing demand for cheap construction methods brought
with it a demand for cheap to build forms. Curves have always been
more complex to imagine and more difficult to build than straight
lines. The appearance of new arches and vaults were common events
in the Renaissance. But, since then the demise of the curve has
been slow and unrelenting. Curves are now the property of engineers.
We see them used in bridges.
I do not propose to show how to build arches and vaults cheaper
than lintels and flat ceilings. I will, however, restate the
structural merits of the curve and show methods for understanding the
variety of curves and their relationship to several kinds of loadings.
In addition, I present an argument for the union of the curve and the
III
straight line in that modest, but ubiquitous container of humanity,
the room; the room with a rectilinear floor plan and a vaulted
ceiling. The wedding of these forms has been tried many times before,
more often in the more distant than the recent past. Now it is
rarely seen, but in a museum or historic building. The juxtaposition
of curved ceiling forms and orthogonal floor plans should be revived
for it was a lively combination and brought cheer to many.
Thesis supervisor:
Title:
Waclaw P. Zalewski
Professor of Structures
IV
Acknowledg ments
I would like to thank Waclaw P. Zalewski, whose inspiration
and encouragement have long been a primary reason and motivation
for pursuing my studies at M.I.T..
To Elizabeth Reed, for her patient editing and sustained
interest in this project.
To Callie Traynor, for setting me rolling.
A special thanks to Kim Ruoff for her assistance in Rotch
Library and for the lending of her necklace for the catenary
drawings.
And to my family, for their support during the undertaking of
yet another adventure.
V
Vi
"Y asi el deber, lo moral, lo inmoral y lo amoral,
la justicia, la caridad, lo europeo, y lo americano,
el dia y la noche, las esposas, las novias y las
amigas, el ejercito y el banco, la bandera y el
oro yanqui o moscovita, el arte abstracto y la
batalla de Caseros pasaban a ser como dientes o
pelos, algo aceptado y fatalmente incorporado,
algo que no se vive ni se analiza porque 'es asi'
y nos integra, completa y robustece."
Julio Cortazar
Rayuela
VII
Vill
Final Design Competition Entry: Fall Term, 1982
Structural Engineering Laboratory 1.105J/4.315J
Student Designers: Garet Wohl and George Schnee
Professor of Structures: Waclaw P. Zalewski
Teaching Assistant: Thomas H. Hartshorne
Ix
X
Contents
1
13
Preface
1. An Introduction
Cu rvature,
Circle
and
43
11.
and
65
Ill. Catenary
97
IV. Vaulted
133
Conclusion
141
Bibliography
145
Illustrations
to
Height/Span
Pressure
Lines
Ellipse
and
Parabola
Ceilings
XI
Ratio,
XII
Preface
Why an arch?
nice.
Arches are beautiful.
Curves in general are quite
The long neck of a swan, the lines of a 1956 Thunderbird, the
S-sense of a woman's body.
Curves in architecture deliver a definition
to space whose form is somehow less definite than those orthogonal
shapes which exist.
A right angle is made up of two lines whose
construction consists of connecting three points with two straight
The conceptualization of a curve
lines which meet at a right angle.
is not so easy.
We needn't restrict ourselves to the shape itself.
What the shape cuts out of the landscape is also of interest.
That which is framed by the arch is more difficult to describe
than a subject framed by a rectangular doorway.
If we are looking
out through a door, the top of the doorway might be coincident with
the belt-coursing of a building across the street.
The jamb might
parallel a drainpipe which frames the edge of a window.
Since most
of the built world heeds the rigor and logic of an orthogonal
ordering, it is often an orthogonal composition that makes up the
field of view framed by the arch.
The arch does not reinforce the
vertical and horizontal lines of the buildings in the background,
but cuts across the face of things showing a corner of a doorway
1
here, excluding the corner of a window there.
So, how can the curve
of the arch be considered beautiful if it often interrupts and moves
in counterpoint to its surroundings?
Leaving the discussion of
aesthetics for later, I'd like to first place before the reader a
rough outline of my intentions in writing this paper.
With this
information he will obtain a better impression of whether or not the
entre'e appeals to his interest.
I will stay on the curve:
at times under, on, and over it, but
consistently on the subject of the curve and its structural and
aesthetic value to architecture.
I'm primarily concerned with how
curves function as effective spanning forms;
floor and roof of buildings.
in bridges, and in the
Throughout the course of the investigation
I will entertain discussions about those curves which can be
categorized under the headings:
circles, ellipses, catenaries,
parabolas, and freehand curves.
In Medieval, Renaissance, and Baroque times arches and vaults
were commonly employed to span openings in walls and to cover space.
Over the last one hundred years, with the advent of steel and the
redevelopment of concrete, they have been used less and less.
thin gage sections of steel are very strong.
Even
Steel is easily
manufactured in linear beam and column lengths, and in a variety of
sections.
Concrete members are economically produced in linear
lengths;
prefabricated or cast in place, pretensioned or post-
tensioned.
Concrete slabs, either with or without steel decking
underneath, are readily formed in planes.
2
Today, in 1983, why use an arch?
With the demand for cheap,
simple, and rapid means of construction, why go with a form that is
not as easily fabricated, connected, supported, or subdivided as
the tried and true horizontal beam?
arches beautiful.
First, most people consider
Also, there is the recognition of their economic
and structural viability for large spans.
In the Gothic Age arched
vaulting created a magnificent overhead landscape, and enabled the
early builders to construct some of the largest spans to date.
Witness, the fan vaulting of King's College Chapel at the University
of Cambridge completed in 1515 (Fig. 1).
Fig. 1
Fan vaulting of King's College Chapel, completed 1515.
3
During the boom of the railroads in England Isambard Brunel
designed and built two wrought iron lenticular arches, 1,000 tons
apiece, whose 465 foot lengths made up the two spans of the crossing
of the Tamar at Saltash.
Fig. 2
Brunel's bridge was completed in 1858 (Fig. 2).
Tamar at Saltash, Isambard Brunel, completed 1858.
In 1932, the 1,652 foot clear span of the two-hinged steel arch
of the Bayonne Bridge in New York State was successfully finished
4
(Fig.
3).
Fig. 3
Bayonne Bridge, New York State, completed 1932.
Aside from its grace as a form and its capability as a long
spanning element, how practical is the arch for the short span, forty
feet or less?
Arch and vault construction tends to be more labor intensive than
post and beam frame or wall and slab construction.
Reasons why have
something to do with the chicken and the egg dilemma.
Understanding
the form and building the formwork necessary to construct curved
5
surfacesis more difficult than that for linear beams and columns.
In modern times with the availability of simpler alternatives, arches
have been built less often.
Due to the infrequency of construction,
the improvements in construction techniques and reduction in costs
have occurred more slowly in this area.
Also, arches and vaults
are less efficient spatially, both in the fabrication and in use.
Prefabrication and in-situ fabrication require methods and formwork
which take up more space.
After they are built they occupy space,
both above and below, which people, machines, furniture, or other
buildings cannot occupy.
Vaulted or domical spaces are unreasonable to subdivide
because of the difficulty of matching partitions to the curvature.
Therefore, these kinds of spaces are rather intractable when the time
comes for remodeling.
One can construct lower height barriers within
a vaulted space, but sound transmission is still a problem.
Placing
ceilings over the barriers denies the aesthetic value of having a
vault.
And with energy costs an extremely important consideration,
one must balance the negative tendency of vaults to act as heat sinks.
If the designer is considering repetitive small span arches,
there is a cost saving due to the repetition of forms, process, etc..
But, there is an additional cost over frame construction because
one must supply the material and structural means to restrain the
arch's outward thrust.
Two arches side by side support each other.
But what of the end bays?
Either they require a significant buttress
or the placement of a tie member from end to end to balance these
6
Given these acoustical, energy, structural, spatial,
thrusts.
and economic difficulties, the arch and vault are still the most
appropriate forms for carrying load over a span.
Their
appropiateness lies in the fact that they develop the full strength
of the material, and therefore can do the job with less overall
material than a beam or slab.
Later, I will go into greater depth
with this comparison.
With these factors in mind, let us proceed to analyze the
mathematical and structural nature of curves.
Even where arches and
vaults are not economically the most feasible solution, their forms
may be the best architectural choice.
with nature.
landscape.
Curves call forth associations
On the outside, they can stimulate and focus the
This is true of the dome.
If well designed, inside a
curve there is a feeling of containment and place as there was in
the womb.
Curves celebrate the challenge of changing slope with
a continuity and grace not found in angles.
They evoke an emotional
response and by contrast, offer a heightened appreciation for their
linear neighbors.
There exists a large variety of curves which are known to us by
their shape and whose dimensions we can compute with fairly simple
mathematical formulas.
these.
Circles, ellipses, and parabolas are among
The more complex catenary curve will also be presented because
of the importance of the loading which generates its funicular shape.
Other curves., such as sinusoidal and curves defined by cubic and
higher order equations, will not be discussed here.
7
However, they
too can be analyzed by the following methods.
The value of the following exercises are manifold.
question:
I ask the
What are the loads which correspond most reasonably to the
above mentioned curves? Designers are rarely asked this question.
Usually, the architect, if he chooses to use an arch or a vault in
his design, selects the curve for aesthetic reasons.
He then gives
the curve, along with the loads specified in the program, to the
engineer.
The engineer's dilemma in this example is the subject of
Chapter IV. If the architect were interested in the structural aspects
of design, he might ask himself:
Given the program loads and the
possibility of using a curved form to support and transfer these
loads, what curve would work most efficiently?
question asked in reverse.
This is my original
At present, I want to skirt program
issues and focus on the nature of curves for the general information
of such an architect in such a situation.
Therefore, I will
approach the problem from the side of the curve.
What does one mean by "structural efficiency?"
First, for both
hanging arcs, and standing arches, there will be as little bending
as possible in the structure.
Loads will be transferred either in
tension, for hanging forms, or in compression, for standing forms.
In this arrangement the stress is uniform across the cross section.
A correct application of the form allows the material of the entire
cross section to be stressed to its allowable limit. As a result we
can engage less material to do the work than in a beam where the
stresses vary enormously.
The stresses (stress =cr = load/area) will
8
therefore be of the axial sort.
The uniform stresses will develop
along the longitudinal axis or "in-line" with the curve of the arch.
Bending stresses also develop in line with this axis, but occur
directly from forces which are "out-of-line" with the curve of the
arch.
The relation between the moments caused by loads and the
internal bending forces which balance them can be explained with the
help of the following analogy.
Find a friend, any friend will do.
representative of the body of the arch.
Consider his body as
To imagine this, think of
his body as having a uniform cross section and consisting of a
single material such as bone.
Have them buckle on a pair of ski boots
and lock into a pair of skis, downhill variety.
us to model a rigid connection to the ground.
their head directly from above.
pressure.
This addition allows
Now, press down on
Try standing on a chair to apply the
You are inducing a longitudinal force in your friend's
body with average stress equal to the force divided by the cross
sectional area of their body.
and straight in front of them.
Have them hold out an arm rigidly
With his body resting in a vertical
position, press lightly down on his wrist (Fig. 4).
You are
introducing a bending moment to the body equal in magnitude to the
applied force times the length of its arm.
If his ankles are steady,
his body will suffer a uniform bending moment down to the ankles.
His body endures a maximum bending stress equal to the bending
moment, M, divided by the section modulus, S, of his body's cross
section:
9
Cr
= M/S
Eqn. 1
The section mudulus is a term which defines the geometric characteristics of the cross section with respect to its ability to resist
bending action.
The curling of your friend's toes as he attempts
to maintain balance suggests that the bending stress is experienced
down to his feet.
The muscles of the feet, legs, and thorax supply
the resisting moment equal and opposite to the applied moment.
It
becomes apparent that friends with larger cross sectional area, and
consequently, larger section
modulus, have an easier time with this
trial.
F.
M a Pa
a
Fig. 4
Moment analogy, ski boots and skis model rigid connection
to the ground.
10
We are better off if the load is applied in a manner which allows
the resultant of forces to be transferred without the added burden
of a bending stress.
The design is more efficient because the arch
can be built thinner and lighter.
Other characteristics which
influence the ability of arches to carry load and withstand stress
will be reviewed in the following section, Chapter I.
11
12
I. An Introduction to Height/Span
Ratio, Curvature, and
Pressure Lines
The strength of an arch depends on the shape and its appropriateness for the load, the modulus of the cross section, S, its substance,
and the help it gets from its structural neighbors.
The latter three
parameters will be explained as part of a design problem in Chapter
III.
Presently, I will introduce several concepts relative to shape
including the importance of the height/span ratio, curvature, and
the nature of pressure lines.
Consider the simple, symmetrical arch with uniform cross section
drawn below (Fig. 1-1).
The stress along the longitudinal axes is
constant throughout:
-=
where:
AT/AA = BT/BA
Eqn. 1-1
AT, BT
= total reactions at the supports
AA, BA
= cross sectional areas of the two legs
If the load, P, is increased the resulting reaction forces, AT and
13
P
C
h
"1=90*
A HB
A.T
A,
BV
BT
L
Fig. 1-1
Symmetrical triangular arch with uniform cross section
under a point load.
P
Where:
load
=
AH, BH
=
horizontal components of the total reactions
A
=
vertical components of the total reactions
BV
BT, will increase proportionally.
The stress in each leg will
increase by:
acr =AA T/A =A
BT/BA
Eqn. 1-2
If the load, P, is kept constant and the angle, o<, is opened from
14
90* to 1350 (either by lengthening L, or shortening h, or both), the
stress in each leg will increase proportionally. The reason for
this change is suggested graphically in the following sketch (Fig. 1-2).
PI
90*
p
T
AH
T
91
2
135*
AH
Fig. 1-2
BH
Two symmetrical arches with uniform cross sections,
but different apex angles.
As the angle, o<, grows from 904 to 1350, the vertical component of
each leg, AV and By, remain at P/2 to equilibrate the load, P. The
horizontal components, AH and BH, of the total reactions, and the
total reactions themselves, AT and BT, increase to absorb the
changing angle.
The magnitude of these reactions is fixed by the
geometry of the arch and the magnitude of the load.
Of course, the
forces, AT and BT, in reality lie in line with the legs of the arch.
15
As cK
approaches 1800, AT and BT become infinite, as do the
resulting stresses.
As c< goes to zero, with the height/span
ratio becoming infinite, the leg forces approach P/2 and the
stresses decrease considerably.
Now, consider a symmetrically curved arch member supporting
a point load, P, at the center span (Fig. 1-3).
P
AT/
AV
B
BT
L
Fig. 1-3
Arch made from segment of a circle, under a point load, P.
For the curved arch the same relationship
triangular arch.
holds as for the
Ignoring stresses due to bending, raising the
16
height/span ratio resolves in less stress along the longitudinal
axis of the arch, provided the nature of the curve is preserved:
circle, parabola, etc..
Obviously, the height/span ratio for a
circular arch cannot be raised above 1/2, because of the circle's
constant radius.
The stresses due to forces in-line with the body
of the arch increase toward the supports.
For a given height/span ratio, a horizontally distributed load
placed over the arch will bring about a stress distribution which
varies from the top to the bottom of the arch (Fig. 1-4).
distributed load
AV
A T/I
I
Fig. 1-4
=
p
BV[\T
L
-
Segmental arch under a horizontally distributed load, P.
17
Ignoring moment stresses, a distributed load will produce the
greatest stresses due to forces in-line with the longitudinal axis at
the supports.
as o<
As the force resultant moves down the body of the arch,
increases, the addition of further loads can only increase
Increasing the
the magnitude of the force resultant and the stress.
height/span ratio for a given curve will diminish throughout the
stresses which develop along the longitudinal axis.
Another approach
to examining the forces which occur in the arch is through an
understanding of curvature.
The radius of
The curvature, K , for a circle is a constant.
curvature, R,, for a circle is equal to the radius of the circle.
Curvature for any curve is equal to the reciprocal of the radius
K. = 1/R.
of curvature:
The radius of curvature at any point
along any curve, is the radius of that circle which has the same
curvature as an infinitesimal segment of the curve which includes
The parabola has a changing curvature.
the point.
As one proceeds
away from the apex, the curve flattens out while the radius of
curvature at each further point increases (Fig. 1-5).
The parabola has the basic form, y
c
=
= ax2 + bx + c. where
the Y-axis intercept and (-b/2a, b2/4a - b2/2a + c) are the
(x,y) coordinates at the apex.
The nominal radius of curvature at
the apex, R0 , is defined by a, the coefficient of x2 in the above
equation, in the following manner:
R
where:
jaj
1/21al
=
=
Eqn. 1-3
the absolute value of a
18
X-axis
Fig. l-5a Parabola with
intersecting line
segment showing relationship of parts.
2
R i = RO/cos 3 xc
Ro= L/8h
= R0 (1 + 2y 1/R 0 )
X - axis
dotted lines
indicate centering
of radii of curvatur
N
N
N
.-
Fig.
y = x2j
1-5b
N
N
N
Parabola showing changing curvature and radii of curvature.
19
Also, R0 has the following relationship to the horizontal span, L, of
any section cut through the parabola, where h is the vertical height
of the section at midspan (Fig. 1-5a):
R
= L2/8h
Eqn. 1-4
RV,the radius of curvature at any point, i, along the parabola, can
be found once R and the angle, o( , which the parabola makes with
a horizontal line at i, are found:
R. = Ro/cos3
Eqn. 1-5
The equations, 1-3, 1-4, and 1-5, hold true for arches, cables,
membranes, and shells with parabolic cross sections.
The height, h,
for an arch, merely becomes the sag, f, for a cable.
The different
methods for determining the radii of curvature at several points
along the curve are important because they assist the designer in
establishing the compression
and tension forces in arches, hanging
cables, and shell structures.
For an arch in compression or a cable in tension we can use
Ri, to find the force, Ti, in the element, if we know the component
of the load perpendicular to the surface, pL:
Ti
= pI R
Eqn. 1-6
Although, I show the circle in the example below (Fig. 1-6), the
relationship holds for any curve.
From these examples it is apparent that a larger curvature, K,
and a smaller radius of curvature, R , (K. = 1/R), will yield a
smaller force in the member or membrane for the same external force,
20
T ten.i =p'R.
TTcomp- =OP
p R.
T
P
Tc
Fig. 1-6
p1
=enp
Tt
R
Tt
Tc
Compression and tension forces in circular membranes.
. Increasing the height/span ratio for an arch under a distributed
load (Fig. 1-4) effectively increases the curvature.
Consequently,
the internal member forces are relatively diminished.
Nature uses curvature in a myriad of instances to protect its
creatures against other forces of nature;
creatures.
such as weather and other
Witness the giant sea clam, (Fig. 1-7), whose corrugated
shell guards the inhabitant against the forces of sea water and
predators.
The greater curvature of a corrugated shell over the
more simple convexly contoured shell, lessens the internal forces
which arise in response to externally applied loads.
21
Fig. 1-7
Corrugated shell of the great sea clam.
Membranes are thin plates in three dimensions whose structural
integrity is maintained primarily by tensile forces.
Shells are thin
plates whose equilibrium is maintained by an interplay of tensile
and compressive forces.
Determining the membrane forces in membranes
and shells is a bit more involved than with the planar configurations
22
such as arches, cables, and rings.
Professor Waclaw Zalewski comments
in his notes on "Membranes and Shells," on the nature of forces in
membranes, "In general two basic conditions determine the forces
within a membrane:
1) curvature and 2) boundary conditions."'
inspect curvature in greater depth.
Let's
The curvature of shells (note:
from this point on I will use the term, "shells," to indicate both
membranes and shells) is characterized as positive, negative, or
zero.
The intersection of the shell with two perpendicular planes
oriented with respect to the principle axes of the shell, delineates
two lines which lie in the surface of the shell.
The two perpendicular
planes are oriented properly if the lines delineated by the intersection with the shell have the greatest difference in curvature of
all possible orientations.
For every shell there is an orientation
of the planes which will describe two lines whose curvatures determine
a maximum and a minimum curvature for the shell.
This is the proper
position of the planes to ascertain the curvature of the shell.
If the curvatures of both lines is oriented in the same direction,
then the curvature of the shell is considered positive.
have positive curvature.
Domical shells
If the curvatures of the two lines are
opposite in orientation, the curvature of the shell is negative.
surface of a riding saddle is exemplary of negative curvature.
The
Zero
curvature is produced by the product of a straight line, with zero
1 Zalewski, W. P. "Membranes and Shells"
p.
23
36
curvature, and a curved line.
A cylinder has zero curvature.
Let T1
be the membrane force in one line and let T2 be the force in the
other line.
The following formula relates the external force
perpendicular to the surface, p1L, and the radii of curvature of
the two lines, RI and R2, to the membrane forces:
Tl/R
Fig. 1-8
+ T2/R 2
p
I
Eqn. 1-7
Surface with negative saddle-like curvature, Eqn 1-7
applies to surfaces with positive curvature as well.
Because the forces at right angles to each other in the negatively
curved surface are in compression and tension, they have opposite
24
signs.
The opposite signature cancels out the effect of the opposite
direction of the curvatures of the principle axes.
terms, Tl/R
and T2/R2, are additive.
radius R =
=
Therefore the
For spherical shells, the
R2. Eqn. 1-7 becomes simply:
Tl/R + T2/R
=
A difficulty with Eqn. 1-7 is the effort required to calculate the
value of the force perpendicular to the surface, p-L.
There are
methods which can assist the calculations necessary to find the
forces in each kind of shell.
I will suggest a few which work for
the hyperbolic paraboloid and various kinds of domes.
For a hyperbolic paraboloid ("hypar") situated under a load
uniformly distributed along the horizontal, the following is true.
Depending on the boundary conditions, the horizontal components, H
1
and H2 , of the membrane forces, T1 and T2 , can be constant throughout.
Imagine the two perpendicular lines in Fig. 1-8 to have a width
equal to one unit of measure.
Because they are situated in the center
of the saddle, a full unit's worth of the distributed load will rest
on these strips.
If we consider parallel strips to either side of
either center strip, their widths are determined by the vertical
projection of a one unit strip of distributed load.
In either
direction, as the slope of the saddle gets steeper the actual strips
will get wider to accept the same one unit strip of load from above.
Since the load acts over some area, giving dimension to the lines
along which the compressive and tensile forces act allows an actual
quantity to be associated with the membrane forces, T and T2'
1
If the hypar consists of similar parabolas which generate the
25
curvatures in the two perpendicular directions, then H1
=
H2 . At
the intersection of the apexes of the two parabolas,
H1
= H2 = T
= T2. The last statement is true because the
shell is horizontal in both directions at this point.
Since the
nominal radius of curvature at the apex, R0 , is the radius of
curvature for both parabolas at this point, we can substitute these
values into Eqn. 1-7:
H/R
+ H/R
2H
= p R0
=
p
Eqn. 1-8
Eqn. 1-9
By definition the load, p, is perpendicular to the surface of the
hypar at the intersection of the apexes.
to:
Eqn. 1-9 can be simplified
2H
= pR .
Since p is a given, by nature of the program and code
specifications, if we know R we can find H. Eqn. 1-3 states:
R
=
1 /21aj
where a is the first coefficient of the standard form
of the parabolic equation:
y = ax2 + bx + c. We need to know
these coefficients to design the curvature of the hypar at the outset.
Also, from Eqn. 1-4 we know:
R
= L2/8h, where L is the horizontal
span of any segment of the parabola we choose and h is the height or
sag of the curve above or below the segment line taken at the midspan
(Fig. 1-5a).
Employing either of the above equations to fix R allows
us to find H. With H computed, we can evaluate TI and T2 at any point
within the boundary or edge conditions.
The edge conditions are too
complex to generalize about.
The horizontal component of the membrane force, H, is constant
26
for the hypar under a uniform distributed load.
At any point along
the shell, the compression force, Ti, equals the horizontal component
of the force, H, divided by the cosine of the angle which the shell
at that point makes with the horizontal:
T
= H/cos o1
Tlmax
Eqn. 1-10
Eqn. 1-11
= H/cosCodlmax.
And for the tensile forces in the shell, the same is true:
T2
Eqn. 1-12
= H/cosoc2
T2max.
= H/coso(2max.
Eqn. 1-13
Fig. 1-9 shows the intersection of a compression strip and a tension
strip of a hypar meeting at some point other than the apex.
The
drawing simulates the elements of the above equations.
If R1 and R2 of Eqn. 1-7 are calculated using Eqn. 1-5
(R
=
R0/cos 3 o ), then T2 can be derived by another method once
H and p-L
are established.
T
=
Rewriting Eqn. 1-7 shows:
(pL - H/coso4 )R2
Rl
Eqn. 1-14
The above methods of analysis also help to understand domical
shells.
The methods, although different in form and result, are
useful if the notion of principle lines of force, indicated above,
is evolved.
For simple analysis of the hypar, one can imagine the
lines to be parabolic strips, one unit measure or greater in width,
whose axes lay parallel to one or the other of the principle axes
of the shell.
With the addition of dimension, the above formulas
gain quantatative value.
The area on which the load, p. rests can
27
T2
H
a2
principle line
of compression
H
T2
T,
pr inciple line
of tension
Fig. 1-9
Intersection of compression strip and tension strip
in a hypar at point other than the apex. Imagine that
the lines shown have dimension. Principle forces and
horizontal components are depicted.
28
be specified to the width of the strip.
Consequently, T1 and T2 can
be derived as the compressive and tensile forces per width of strip.
For stress, one would have to consider the thickness of the shell as
well.
For the dome, the strips are best taken in a manner whereby
the strips radiate from the north pole or highest point.
are referred to as meridian strips.
These strips
The forces which develop in-line
with the strips are called the meridian forces.
The membrane forces
which evolve in the direction prependicular to the meridian forces
are referred to as the ring forces.
The meridian forces are generally
compressive while the ring forces shift from compressive forces
towards the top to tensile forces towards the bottom.
Once the
meridian forces, Ti, are found, the ring forces, T2 , can be figured
with the form variation of Eqn. 1-7 written below:
T2
=
(p
- Tl/R )R2
Eqn. 1-15
With this brief survey of shells and the workings of the
internal forces under load, I will now return to investigate more
thoroughly thinner slices of structure;
the nearly two-dimensional
arch.
What are the necessary conditions for an arch to be placed in
compression and how does one define the limit at which bending forces
begin to exist?
If the force resulting from the addition of loads
to the arch above the point of analysis is located within the inner
1/3rd of the arch's depth and within the inner 1/3rd of the arch's
width and oriented such that it is parallel to the central axis of
the arch, then the entire cross section of the arch will be in
29
compression (Fig. 1-10).
R
d
uniform average stress
over entire cross sectional
area: crav = R/bd
Fig.
1-10
Cross section of arch member showing inner core within
which the force resultant can place the arch in
compression. Where:
R
=
force resultant
d
=
depth of the arch member
b
=
width of the arch member
CA
=
central axis
If the resultant of the load forces, R, is directly in-line
with the central axis (as drawn above) then the average stress
experienced by the arch is uniform across the cross section and is
30
equal to the force resultant divided by the cross sectional area,
Crav
R/bd.
=
Any eccentricity, e, created by the load resultant's
position occurring at some distance away from the central axis,
produces a moment stress.
~max.
=
The maximum stress due to bending,
M/S, is therefore defined by the moment, M = Re, and
the section modulus, S = bd2 /6:
a- max.
=
Eqn. 1-16
Re
bd2 /6
A linear elastic material is a material which when loaded experiences
stress (0') values in direct proportion to strain (E) values over
a certain range (i.e. where:
E = (-/E ). Moment invites the
following stress distribution in a member with linear elastic
properties (Fig. 1-11).
M = Re
R
t
0-tensile
compressive
CA
Fig. 1-11
Stress distribution due to moment, M
linear elastic material.
31
=
Re, in a
Once the magnitude of the maximum tensile stress due to bending
equals the compressive stress due to the force acting in-line with
the longitudinal axis, then the limit of zero stress has been reached
in the extreme fiber on the tension side of the member.
The stress
is zero at this point because the tensile action cancels the
compressive -action.
compressive axial
R/bd
=
G tensile bending
Re
Eqn. 1-17
Eqn. 1-18
bd2 /6
e
=
d/6
Eqn. 1-19
As depicted graphically in Fig. 1-10, this limit occurs at a distance
of d/6 to either side of the central axis.
the inner 1/3rd
If R is located within
of the arch's depth and within the inner 1/3rd of
the arch's width, the entire cross section will be under compressive
stress.
Once the limit is surpassed, the extreme fibers of the edge,
farthest away from the force resultant, are placed in tension
(Fig. 1-12).
In Fig. 1-12b, with e
=
d/2, the inner edge has reached four
times the average compressive stress,
0 -av,
that would arise if R
were located directly above the central axis.
The outer edge is in
tension, a force not all arches are able to resist.
If the eccentricity
is greater, the resulting moment and bending stress will grow
accordingly.
If we can trace the trajectory of the force resultant
and its relative position to the arch we can determine the overall
moment stress distribution.
32
M= Re
al limit state:
e
=
d/6
R
bi
-
tension state:
e =d/2
e
Cav
(Co m
(co m p.)
CA
(oa v
2 crav
(COMP.)
(t en s.'
CA
crav = R/bd
Fig.
a)
1-12
Limit state:
-
b)
Arch in tension:
Where the total stress in the arch develops
from the force resultant, R, acting at an
eccentricity, e = d/6.
Total stress shown in arch where the stress
arises from the force resultant, R, acting
at an eccentricity, e = d/2.
33
R
For arches with prismatic cross sections whose material behaves
in a linear elastic fashion (e.g. steel), within the linear range
the moment stress will manifest itself in a linear distribution
from maximum compression on the side of the arch toward the eccentric
loading to a minimum compressive or maximum tensile stress on the
Take for example the loading of the arch in Fig. 1-3:
opposite side.
a point load, P, at center span, over a curved arch.
For the purposes
of this example imagine the arch to be a semicircle.
The pressure
line, indicates the optimum path that the load could take to the
supports.
For any arch the pressure line is the funicular shape
under a similar load, flipped upside down.
The pressure line for
a point load is made up of two straight lines which connect the load
and the supports (Fig. 1-13).
The area enclosed by the pressure
line and the ground plane is approximately equal to the area enclosed
by the arch and the ground plane.
To find the height of the triangular
shaped pressure line, the area of the triangle is equated to the area
of the semicdircle:
hL/2
where:
L/2
=
=
1T(L/2) 2
2
Eqn. 1-20
R = radius of the circle
Eqnl 1-21
h = .785L
The moment, at any point on the arch, caused by the effect of
the out-of-line pressure line is equal to the horizontal component
AH or BH, times the vertical eccentricity, e
,
at that point.
The
vertical component of the pressure line force, AV or BV, passes
through the point at which the moment is taken.
34
It therefore has
P
al Semicircular arch
pressure line
pressure
I
and
I-
line--'
h
R
A
/
A/A
Br\T
ATAV
L/2
L
bi Moment
diagram
Ma x. pos. moment
.1PL
g
(09PL)
9P
.05PL
Max. neg
- .o5PL
Fig.
1-13
moment
(-.04PL)
a)
b)
Development of the pressure line for a point load
placed at midspan over a semicircular arch.
The resulting moment diagram.
35
zero moment arm and zero moment.
In this case the slope of the
pressure line is constant as is the force throughout
each leg, AT
and BT, Consequently, AH and BH are constant as well:
AH
= BH
= .32P. At the apex of the arch, the maximum positive
moment is found with a vertical eccentricity equal to .285L:
MC
=
.32P(.285L) = .09PL
Eqn. 1-22
The moment goes to zero as e goes to zero at the approximate points,
L/4 and 3L/4 (Fig. 1-13b).
The maximum negative moments occur at
the approximate points, L/8 and 7L/8, where:
M =
.32P(-.125L) = -.04PL
Eqn. 1-23
To counteract the increases in the magnitude of the moment, the
cross section can be thickened, more reinforcing bar can be added,
or both.
These changes would resist the tendency of the load to fold
the arch inwards under positive moment, or outwards under negative
moment.
Every arch has a geometry and every corresponding load has
a trajectory which describes the pressure line relative to the arch.
The graphic simulation of the interplay of load forces and arch
geometries affords a means for quickly estimating the necessary
reinforcing and cross sectional area throughout the length of an arch.
Obviously, if the geometry of the pressure line and the arch coincide,
we have a more efficient system.
The beam, like the arch, is a mechanism for transferring load.
And, like the arch, the beam sustains the weight of a load by
developing resistance to compressive and tensile stresses.
For any
load, both the arch and the beam will transfer the load forces through
36
the material of the member to the supports along lines which are
called trajectories of principle stress.
The principle stress lines
for tensile stress lie perpendicular to the principle stress lines
for compressive stress.
For a uniformly distributed horizontal load,
the principle lines of stress for a rectilinear beam and an arch
of uniform cross section are shown below (Fig. 1-14).
Assume the
arch is the inversion of the funicular shape for this load.
The principle lines of stress in the beam occur in a concentric
arch-like and cable-like configuration.
These curved lines have a
shape somewhat similar to the pressure line for the load.
Although
the stress patterns which are present in the arch and beam manifest
themselves in similar ways, the differences between the two are
significant.
The arch develops compressive stress along the longitudinal axis.
The average compressive stress is uniform throughout the cross
section because in this case the pressure line coincides with the
geometry of the arch (both are parabolic).
The dotted lines indicate
the lines along which principle tensile stress would arise if it
existed.
But, since there is no force present to generate tension,
these lines merely represent the axes of zero compression.
In the beam maximum compressive stress takes place at the top
center while the maximum tensile stress lies at the bottom center
(Fig. 1-14).
Elsewhere, these stresses diminish in intensity.
Because the beam will attain full stress capacity in only a few spots,
there is a less than efficient use of the material.
37
As a result, the
uniform
load
LLL L
L
L L L LI
region of max. compressive stress
lines of
principle
tensile
------
..
-/
stress
/
--
1
region of max.
tensile stress
--
/
r
lines of
principle
compressive
stress
LIL L LLL LL L LL LL L L
dotted lines:
idicate lines of
minimum stress
lines of principle
compressive stress
region of
toressive
/
Fig.
1-14
Beam and arch under uniform load showing principle lines
of compressive and tensile stress,
beam will require considerably more volume of material to do the
38
same job.
A drawback with the arch is that it requires a tie member
or a buttress to resist the horizontal thrust.
Its height at the
center span may very well exceed the depth of the beam to advantage
or disadvantage.
If the height of the arch were equal to the depth
of the beam the thrust forces would be large indeed.
A concept emerges from the above discussion that is significant
enough to reiterate here.
When analyzing the load bearing capability
of structural members,the appropriateness of the geometry of the
section
for the given loading
is the most critical factor in
assessing its structural efficiency.
Again, structural efficiency
suggests the potential of the form to develop the full strength of
the material.
If one cuts a section through both the arch and the
beam of Fig. 1-14, the point is graphically demonstrated (Fig. 1-15).
The external forces of the load, P, and the vertical support
reaction, AV, will try to rotate the member and turn it over.
The
beam has an internal moment arm within its section equal to approximately 2d/3 to resist the overturning.
The internal couple of forces
need to be quite large in comparison to the external couple of forces,
whose moment arm is much greater.
The arch, however, resists the
overturning with the help of the horizontal thrust, AH and the other
half of the arch.
The moment arm of these resisting couple of
horizontal forces is generally of a magnitude similar to the moment
arm of the vertical couple of forces.
Therefore the internal forces
which arise to counteract the load on the arch are relatively small.
In addition the moment due to vertical forces is absorbed primarily
39
a) Beam
half section
d
E
5
P
2d/3
M = F(2d/3)
Avi
L/2
P
b) Arch half section
H
M =Hh
h
AH
AVi
L/2
Fig.
1-15
a)
b)
Section of a beam showing the internal couple
of horizontal forces which resist overturning.
Section of an arch showing the couple of horizontal
forces which resist overturning.
40
by the resisting moment due to the arch's geometry and not that
which arises from the depth of the arch.
Again, if the arch's
geometry is such that the load forces can be transferred without
the added burden of a bending stress, we are a step ahead.
can be thinner and lighter.
The arch
With these insights in mind, I will
sketch in the characteristics of the semicircle and the semiellipse
and their orientation to load in the following section, Chapter II.
41
42
Circle and Ellipse
11.
The condition of no bending in an arch is not always possible
with the erratic and unpredictable nature of live loads.
Consequently,
structures have to be designed to resist the eccentric loadings and
resulting bending stresses caused by inhabitants, wind, snow, etc..
However, structures can be designed so that the dead load is supported
with virtually no bending stress.
The latter condition is that which
I would like to derive for the several curves previously discussed:
the best distribution of the theoretical dead loads for a given arch
shape.
Although we will be working in two dimensions, the results
are similar in three dimensions.
Simply imagine each arch and its
distributed load rotated about the axis of symmetry to supply the
information pertinent to the related domical shape.
Let us begin
with the circle (Fig. 2-1).
The circle, with radius R, is centered about the X-axis and
the Y-axis, both measured from the origin, (0, 0).
From the
Pythagorean Theorem we know the following:
x2 + y 2 =
y
=
2
Eqn. 2-1
(R2 _ 2)i
Eqn. 2-2
The latter equation is a description of the circle in terms of y.
Given that R is a constant, equal to any value of our choosing, we
43
Y-axis
X2 + Y = R
(x, y)
y
X-axis
(0, 0)
x
Fig. 2-1
Circle with radius, R.
can plug in values of x and obtain the corresponding y values.
abscissa, x, can be positive or negative.
The
For the purposes of
depicting a semicircular arch, positive values of the ordinate, y,
will be used.
For all statically designed structures, the following three
conditions of equilibrium hold:
44
a)
I FH
= 0
b) IF
=0
Translated these read:
0
.2M =
C)
Eqn.
2-3
a) Summation of forces horizontal equals zero,
b) Summation of forces vertical equals zero, and c) Summation of
moments equals zero.
The forces indicated by these notations include
all forces, both external and internal.
The relationships are true
for any point on the structure, any free body section, and for the
structure as a whole.
To understand how these conditions of equili-
brium work, an explanation will be made with the semicircle as the
example.
Imagine a cable loaded such that the funicular shape that
it adjusts to under the loading is that of a semicircle (Fig. 2-2).
A
0,0
X-axis
B
AH1
A bl
H
By
cable
Cl
x, y
Y-axis
distributed load
I
Fig.
2-2
-L
L
L
V
_
'U,
Semicircular cable under unknown distributed load.
45
In Fig. 2-2
I have drawn the distributed load in a random fashion
because I'm assuming we still do not know the correct one.
AV and
By are vertical reactions at the supports, while AH and BH are the
horizontal reactions at the supports.
C is any point on the cable
with the corresponding coordinates, (x, y).
interaction
At point C, we have the
diagramed below in Fig. 2-3a.
RT
LH
L
Vj e
RV
C
La
load
LT
0(
(x
L.
L
RH
load
load
Fig. 2-3a
where:
Fig. 2-3b
Vector analysis of
point C on cable.
LT
T
Vector analysis
of two different
points. Right
drawing indicates
point closer to
support point B.
total tension left
RT
=
total tension right
=
right vertical
component
LV
=
left vertical
component
RV
LH
=
left horizontal
component
RH
right horizontal
component
For point C to be in equilibrium, the first two conditions show
46
that the following relationships must hold:
I FH =
2F
0:
= 0:
LH + RH
Eqn. 2-4
= 0
LV + R + load
=
0
Eqn. 2-5
The horizontal forces, LH and RH, must be equal and opposite.
This
is true for any point along the cable if the load is vertical.
Eqn. 2-5 states that together, LV and the load, must be equilibrated
by RV, as shown in Fig. 5a.
Another way to understand this relation-
ship is to examine the interaction between LT and RT as we ascend the
cable toward the right support.
The resultant of the loads applied to the left of C is the force
LT." The combination of LT and the additional load applied at point
C must be equilibrated by RT.
As we approach the support point B,
LT increases in magnitude as more load is taken on.
Since the
curvature of the circle is constant, as the resultant force grows,
a greater and greater load will be required to rotate its inclination
the same number of degrees, c>.
Fig. 2-3b.
This phenomenon is diagramed in
Keep in mind that each vector represents position,
magnitude, and direction.
A more difficult interaction to comprehend is the equilibrium
of moments.
Taking moments about C to calculate the resultant
moment acting at C, we can isolate the section to the right of C
with a free body diagram (Fig. 2-4).
To take the moments about C we
must satisfy the third condition of equilibrium:
2 MC
0.
Throughout this paper I will consistently use a sign convention
whereby counterclockwise rotations are considered positive, and
47
R - x
(0,0)
Y -axisf
,V
dist. load
s
R
Fig. 2-4
where:
Free body diagram of the section of cable to the right
of point C.
RF
=
resultant force of the distributed load
MC
=
internal moment at C
s
=
horizontal distance from point C to force
RF'
clockwise rotations are negative.
+, IMC
=
Therefore:
-RFs + BV(R - x) - BHy + MC=
0
Eqn. 2-6
Since the flexible cable, unlike your friend's shoulder, is
incapable of developing an internal resisting moment, MC in Eqn. 2-6
is zero.
This leaves:
48
2 MC
=
-RFs + Bv (R - x) - BHy
=
0
Eqn. 2-8
-RFs + Bv(R - x) = BHy
Both RF and B act in the vertical direction.
horizontally.
Eqn. 2-7
Only BH acts
Thus:
2 Moments due to
vertical forces
=
-2
Moments due to
horizontal forces
Eqn. 2-9
This condition is always true for the hanging cable because of its
inability to sustain internal moment.
With a change of load, the
cable adjusts its shape so that the condition still remains true.
However, this interrelation is not always true for the arch.
Arches
have the capability to develop some internal resistance to bending.
They remain standing because of this ability.
But, if an arch has
the inverted shape of a cable under the same load, it will have no
need to develop bending resistance.
A simply supported beam does not require a horizontal force to
maintain its equilibrium.
The moment, M , taken at any point x,
suffered under loading is equal to a summation of moments due to
vertical forces only.
For an identical vertically distributed load,
the beam is a cousin of the cable, for the moments due to vertical
forces are the same for both.
Consequently, the moment for the beam,
M , is also equal to the summation of moments due to horizontal forces
for the cable system:
M
=
(beam)
2Moments due to
vertical forces
(both systems)
M
=
-2
= Hy
Moments due to
horizontal forces
(cable system)
Eqn. 2-10
Eqn. 2-11
49
Mx/H
=
Eqn. 2-12
y
H = horizontal thrust
where:
y = vertical displacement of cable
For a given vertical loading, H is constant at every point along
the cable.
As a result, MX is directly proportional to y. The
moment diagram of a simply supported beam has the same shape as the
funicular curve, when both are loaded similarly.
Remember, M
describes both the summation of moments for any point on a beam and
the summation of moments due to vertical forces for any point along
the cable.
Now let's return to the semicircular arch.
There are two equations which help us to define the arch:
y = (R2 _ x2)a
Eqn. 2-2
y = Mx/H
Eqn. 2-12
The first describes the semicircle analytically.
The latter equation
is applicable to any hanging cable and to those standing arches where
there exists no internal moment.
If there existed an internal
moment, the moments due to vertical forces would not cancel the
moments due to horizontal forces, and Eqn. 2-12 would not hold true.
The y value in this equation refers to the vertical distance from
the horizontal plane containing the supports (X-axis) to the cable
or arch.
Imagine the vertically distributed load of the arch as first
passing through a horizontal plane superimposed above the arch, before
actually reaching the arch.
In this light the load appears positioned
similar to an identical load applied to a simply supported beam.
50
At
present, we are only interested in the dead loads, basically, the
weight of the structure itself.
If the above two equations are
combined we find:
Mx/H
=
y
=
=
Hy
=
H(R 2 _
=
(R2 _
2)
Eqn. 2-13
H(R2 _ 2)
Eqn. 2-14
Eqn. 2-15
2
The last of these equations states that the vertical ordinate,
y, of a semicircular arch, at any point x, when multiplied by the
horizontal thrust gives the value of the moment due to vertical
forces, M . To obtain the loading distribution for the semicircular
arch the second derivative of the moment due to vertical forces, M
needs to be found:
M '
where:
Eqn. 2-16
= -p,
''
indicates the second derivative
p
= load per linear unit of distance
The value of p is negative because the load acts downward in the
negative direction.
The first derivative of M will yield the
value of shear, V , which is equivalent to the slope of the moment
diagram of a similarly loaded beam:
Mx'
where:
= Vx
Eqn. 2-17
indicates the first derivative
V
= value of the shear force at any point x
Shear is the transverse tearing force experienced by a structural
member due to the opposition of the upward reaction forces and the
51
Find the equation for the shear force:
downward load forces.
M
H(R2
=
M '
x2)e
V
=
Eqn. 2-15
_ -
(i(R
=
Eqn
(-2x))
. 2-18
The derivation gives the value of shear at any point x distance from
the origin.
H is constant and does not change throughout both
Finding the slope of the shear diagram yields the load
derivations.
p at any point x:
V''=
H(-x 2 (R2
p
p
H(-L(R 2 _
=
=
-
2
-
-
(R
+
= -H( x23
-2
1
(R2
-
-
x 2 Y-)
Eqn.
Eqn . 2-20
)
x
2-19
)
Eqn. 2-21
2
x 2)z
2
(R
(R2 _
p
-(-2x)(-2x)
+
2
-H(
2
Eqn.
1I)
2-22
y
3
p
Eqn. 2-23
-H( x2 +y
y
2
-H( R 3
=
px
Eqn. 2-24
y3
Since the load acts downward, p , the load per unit length, accepts
a negative sign:
-p
p
=
=
-H( R2
3
y
2
H( R )
3
)
Eqn. 2-25
Eqn. 2-26
y
Eqn. 2-26 informs us that p varies inversely as the cube of y,
the vertical distance between the ground plane and the arch.
central axis where y
p0
=
At the
R, p is relatively small:
Eqn. 2-27
= H/R
52
Substituting p0 for H/R into Eqn. 2-26, we find:
p
= PO (R/y) 3
Eqn. 2-28
As y approaches 0, at either abutment, p goes to infinity.
The
theoretical dead load distribution for a semicircular arch, radius R,
with axial compressive stress and no bending stress is shown below.
PR =
Px = P0 (R/y)
3
load p
PO= H/R
(X, y)
AV W(0,
0)
X-axis
y =(R 2 X2 )/2
Fig.
2-5
Semicircular arch with the theoretical dead load distribution for axial compressive stress and no bending stress.
53
An infinite load at the abutments and infinite vertical support
reactions to counterbalance are hardly realistic possibilities.
Nonetheless, the diagram offers a general indication of the load
distribution.
The above loading diagram presents a semicircular arch
loaded in a manner somewhat similar to the way the Roman
were loaded.
aqueducts
The Roman loading would have been even more appropriate
if they had filled the hollows behind the spandrels with rubble.
Fig. 2-6
Roman acqueduct in Segovia, Spain.
54
In his book, Cable Structures, Max Irvine comments that the Romans
did just that,
"The Roman arch almost invariably consisted of a
semicircular ring of voussoirs backfilled to provide a level upper
surface."2
The semicircular arch might have generated considerable respect
from the Romans because of how well it behaved in this application.
Not totally understanding the efficacy of alternative loadings, the
Romans used the semicircle in other situations such as the dome,
where it failed to perform as well.
The mason's work in constructing
the Roman arch was simplified by the fact that each voussoir was
cut to an identical taper.
to build with that form.
This time saving reality was reason enough
In addition the Romans may have felt an
intellectual and spiritual link with the perfect unity expressed
by the circle.
Both in plan and in section the semicircle suggests
the whole figure,
the circle;
whose closed form signifies continuity,
and whose curvature represents constancy.
The form of the
aqueduct is often found in nature.
A March
snow storm in Boston left the snow seen in the photo below on a
fire escape bridge in Somerville (Fig. 2-7).
weather melted away the snow.
The ensuing warmer
Warm air passing up through the cracks
between the planks of the bridge precipitated the melting from
underneath.
As the snow melted from above,the water runoff helped
to consolidate the snow below.
The snow that became most compacted
2 Irvine Max Cable Structures (Cambridge, 1981) M.I.T.
Press, p. 11
55
Fig 2.-7
Snow bridge on fire escape in Somerville.
was that which lay in line with the maximum compressive pressure
lines created by the over layer of snow.
Because of the air's
ability to penetrate, the remaining less compacted snow under the
arches melted away.
The use of pressure lines in arch design will
be demonstrated in Chapter III.
With the example of the circle in mind, we will now look at a
very similar kind of curve, the ellipse.
The curve which defines
its form falls between two circles of different radii, a large circle
with a radius a, and a small circle with a radius b. In the following
56
example the ellipse intersects the large circle at the X-axis and
intersects the small circle at the Y-axis (Fig. 2-8).
Fig. 2-8
An ellipse circumscribed by the circles which define it.
57
The ellipse has two foci, whose distance from the Y-axis is c.
If one attaches either end of a string to points C and CR with
enough slack so that it can be stretched taut to make contact with
a third point, BT, a pencil can be inserted on the inside edge of
the string and rotated about the foci to draw the ellipse which
corresponds to the two circles.
How to establish points CL and CR
It so happens that the length of the string necessary
initially?
to draw the ellipse equals 2a.
If we inspect the triangle
CL - CR - BT, formed when the pencil is at BT, we find two right
angles back to back (dotted line in Fig. 2-8).
The hypotenuse of
each is equal to half of our string length, or a. The vertical leg
of both triangles is b, and the remaining leg is c. Therefore:
c
c
=
b2
a
=
Eqn. 2-29
(a2 - b2 )2
Eqn. 2-30
The circle of the first example was expressed in terms of x, y,
and R:
x2 + y 2
x2 +
R2
=R2
=
1
Eqn.
2-1
Eqn.
2-31
R2
The ellipse can be similarly expressed by substituting the radius
values, a and b of the two defining circles, in place of R:
2-32
2
a
2
a2 b22
y = (b2 - b2x2)
2
a
Eqn. 2-33
Eqn. 2-33 denotes the geometry of the ellipse.
58
As with the
circle,
if the value of y, at any point x, is multiplied by the
horizontal thrust, H, we obtain the momeint due to horizontal forces.
This moment value is equal to the moment, M , due to the vertical
forces of load and vertical support reactions:
M
= Hy
2-11
As stated previously, the above relationship is always true for the
hanging cable, and true for the standing arch only when there is
no development of internal resisting moment.
The latter condition
is the dead load design criteria for these case studies.
Again, the
moment due to the vertical forces, M is also equal to the moment
value for a simply supported beam, with a length equal to the span
of the arch, loaded with a similar distributed load.
This is true
because the vertical support reactions for the beam are the same
as those for the arch.
Combining Eqn. 2-33 and Eqn. 2-11 establishes
the following relationship:
Mx/H
=
y = (b2
M = H(b2
-
-
2 2)i
Eqn. 2-34
b2x2 )
Eqn. 2-35
a2
Taking the first derivative of the moment equation will give
the value of shear, V
M
=
V
at any point x along the X-axis:
=
H(i(b2 - b2 2 - i((-2b 2x))
a2
Eqn. 2-36
a2
The second derivative of the moment equation gives the value of the
load, p , at any point x along the X-axis:
59
M'
= p
= H(-(b
x
=-2
2
- b2 x2 )- (-2b2 x)(-2b2 x) + (-2b 2)(b
2
2
2
a
a
a
a
-
b2 x2
2
a
Eqn. 2-37
= H(-b 4x2(b2
p
2
-
a4
(b2
a2
a2
Eqn. 2-38
b2
a2
let g = x/a
p=
2)~
gb
x 2((1 _- g
pH(-b
-
a2
a2
p=
-H(b_(g
2
=
Eqn. 2-40
+ 0 - 92
a2
p
Eqn. 2-39
_a 2 (1 -g 2-i)
-g 2 )1
(
Ecn. 2-41
-Hb
a2 (1 - g2 )g
Again, since the distributed load acts downward, M
2 Hb
=
-p :
Eqn. 2-42
-Hb
-p
X
=
2)
Eqn. 2-43
a
g
)-
Substituting x2 /a2 back in for g2 :
PX p =
Eqn.
2( Hb 2-
ag
2-44
) - x/a
And, substituting y3/b3 in for the (1 - x2
2az
term,
Eqn. 2-44
reduces to:
p
=
Hb4
2 3
ay
Eqn. 2-45
Like the circle, if x = 0, (where y = b), the value of p is
relatively small:
p0
=
Eqn. 2-46
Hb
2
60
Replacing Hb/a 2 with p in Eqn. 2-45, we obtain:
0
p
=
PO (b/y)3
Eqn. 2-47
The load, p , varies inversely as the cube of the ordinate, y. If
x = ±a, where y goes to 0, p becomes infinity.
are also similar to the circle.
These characteristics
For both the semicircular and semi-
elliptical arches, as the sides of the arch approach the vertical
plane, the loads must increase dramatically to turn the resultant
force straight down (Fig. 2-9).
P3
central
axis
1
R2
CK2
R2'1
R
3
c<1
Fig. 2-9a)
b)
W
2
Diagram of the relationship of the distributed load
and the force resultants for a semicircular arch.
Force vector diagrams of two different points on
the arch.
61
p2
where:
note:
p
=
distributed load
R
=
force resultant representing the addition of
previous force resultants and loads
for curvature of the circle to remain constant,
M
=
In the above diagram the magnitude of the load, p, is determined
by the force which is necessary to maintain the force resultant
of previous loads pointing in line with the body of the arch in a
position close to the central axis.
The direction and position
insure that the arch will remain in compression.
The distributed
dead load that would produce compressive stress with no bending
stress for the elliptical arch is shown in Fig. 2-10.
From Eqn. 2-45 it is clear that the load, p , varies directly
with the horizontal thrust, H. By intuition we would expect a rise
in the horizontal thrust to compensate an increase in the distributed
load.
But, as the arch becomes vertical toward the supports and the
resultant thrust approaches the vertical, the support reaction
needed to equilibrate it also approaches the vertical in the opposite
direction.
If this is true, what need is there for a horizontal
component to the support reaction?
thrust at the supports;
throughout?
What happens to the horizontal
a component that is supposedly constant
Since the load is undefined at the limit where y goes
to zero, H maintains its constant value, but with respect to an
infinite value of the shear, Vx, the value of H is negligible.
With this introduction to the process of defining the most
efficient dead load distribution for curved spanning structures,
62
2
p a=4,-0
P-a = 00a
px = Po (b/y)3
load p
Po = Hb/a2
| |
|
|
|
|
|
|
|
|
|
Y-axis
------
Y
H
xis
(0,0)
ALV -o
00
y = (b
Fig.
bX2 )/2
a2
Diagram of the theoretical distributed dead load
for a semielliptical arch.
2-10
where:
-
=
H
=
horizontal thrust
ALV and ARV
=
vertical reactions at the supports
63
ARV[
we are now ready to discuss the effects of live loads.
In
Chapter III, with the aid of the catenary and the parabola, I will
outline a method for designing an arch structure to sustain the
various kinds of live loads to which it is subjected.
First, I
will briefly summarize the dead load distributions which work most
efficiently when combined with these curves.
Next, a graphic
comparison of the arch shapes and the corresponding dead loads for
the circle, ellipse, catenary, and the parabola will be presented.
A design problem encompassing dead and live load factors for a
barrel vaulted sports
arena will conclude Chapter III.
64
Ill.
Catenary and Parabola
A catenary is the curve formed by a completely flexible cable
hung between two fixed points.
The word catenary is derived from
the latin word, catena, meaning chain.
In terms of loading, the
funicular shape taken by a cable loaded uniformly by its own weight
along its length is that of the catenary.
The catenary is the
logical form to support the weight of the arch itself, since naturally
the weight of the arch is distributed uniformly along its length.
Supporting the dead load of a structure is often the single most
important consideration.
Therefore, the catenary is a significant
shape in structural design.
Observe the symmetrical arrangement
of a catenary in Fig. 3-1.
The engineer Prem Krishna in his book, Cable-Suspended Roofs,
derived the expression for the interrelation of the length of the
cable, L, the load, q per unit length of the cable, and the vertical
projection of q onto a horizontal line, px :3
p
Eqn. 3-1
= -q(dL/d )
3 Krishna, Prem Cable-Suspended Roofs
McGraw-Hill Inc. p.30
65
(New York, 1978)
\A
A
(0, 0)
T/ X-axis
A
B
AH
AA 1
[B
BVB
Load
C
q/unit length
of cable
Y-axis
Fig 3-1
Symmetrical cable loaded uniformly along its length
to form a catenary.
where:
dL
=
differential change in the length of the cable
d
=
differential change in the horizontal component of
the slope of the cable
When we examine the horizontal load distribution which gives
the catenary its form we find that the load, p , is equal to the
suspended load, q, at the middle of the span;
one.
where dL/dx becomes
At this point the slope of the cable is zero.
Towards either
support, p,, increases with the gradual increase of dL/d . dL/dx varies
as a function of the angle,
CK.
If cK
= 450 at the supports,
the load, pX, Just inside the supports equals 1.4q.
The distribution
also holds for the standing arch with no bending moment, where q
66
represents the load per unit length of the arch.
The distribution
of p , the vertical projection of q, is shown suspended over a
catenary arch in Fig. 3-2 below.
p
- -q(dL/d
)
po = -q
I I I
I j I
I
I
I ~ I
I j I
II ~ I
Y-axis
I
I
I i I
dX
I I
I I
AH
I
BH
-
(0,0)
X-axis
AV
Fig. 3-2
BV[
Diagram of distributed dead load for a catenary arch.
The load variation is nowhere near as extreme as either the
semicircular or semielliptical arches.
Having approached a more
uniform load distribution with the catenary, the next step would
67
be to ask what funicular form is generated by the uniform load?
To secure the result the process that has been used until now
To begin, I restate the equation of the
needs to be reversed.
second derivative of the ordinate, y'', this time with a constant
load, px, and the horizontal thrust, H:
y''
=
-p /H. Integrating
twice yields the moment due to vertical forces, M , divided by the
horizontal thrust, H (Eqn. 2-12):
p
= -
JY''d
=
y
=
M /H
Eqn. 3-2
Dividing M by the horizontal thrust, H, defines the ordinate, y,
at any point x along the X-axis.
the geometric form.
Knowing y enables us to describe
The first integral of Hy'' gives the shear
force, V :
xx
Hy''dx
0
where:
=
Jpdx
=
V
Eqn. 3-3
0
x = variable which operates between the limits 0 and L.
V
=
Eqn. 3-4
-p + C1
To find C1 we must examine the boundary conditions and substitute.
We know that for a uniform load the shear force will be 0 at
x = 0, at the apex of the arch.
Therefore:
V
= 0 -p(0) + C1
Eqn. 3-5
C1
= 0
Eqn. 3-6
Vx
= -px
Eqn. 3-7
Now, integrating the shear equation yields the moment equation:
68
x
x
J Vd
M
=
-
Eqn. 3-8
M
{fpxdx
= -px2/2 + C2
Eqn. 3-9
The moment due to the vertical load becomes 0 when y
x
=
0, and
± L/2:
=
M
=
0
=
-p(L/2) 2 + C2
Eqn. 3-10
2
Eqn. 3-11
pL 2/8
C2
=
M
= -px2/2 + pL2/8
Eqn. 3-12
This is the equation of a parabola with Y-axis intercept, pL2/8.
Parabolas have the general form:
y = ax2 + bx
where:
+ c
Eqn. 3-13
c = Y-axis intercept
x = -b/2a where slope equals zero at the apex
In Eqn. 3-12 above, b = 0. Another way to think of uniform
load and its relationship
to the parabola is to imagine the effect
that gravity has on a ball as it moves through space.
If you throw
a ball into the air with some amount of horizontal force, the curve
it traces as it ascends and descends is that of the parabola.
The parabola is the funicular shape that a cable will assume
under a horizontally uniform load.
For the arch it is the shape
that will allow a horizontally uniform load to be carried with no
bending moment.
y = MX/H.
This is the case which satisfies the equation:
Any parabola with the character of Eqn. 3-13 will
satisfy these requirements.
By manipulating the parameters a, b,
69
and c, we can adjust the curvature of the parabola and determine its
location relative to the X and Y axes.
The curvature refers to the
sharpness of the curve at any particular point.
Once three points
are fixed the nature of the parabola is determined.
By setting the
span and the height of the arch (or sag of the cable) a specific
parabola with a given curvature is defined (Fig. 1-5).
The location
of the X and Y axes is arbitrary and a matter of convenience.
By establishing the span, L, the height, h, and setting the
position of the parabola with regards to the axes, we fix the values
of a, b, and c. This enables us to draw the arch.
is to plug values of x into the equation,
ordinates of the arch.
Another method
y = M /H, to find the
The latter method is useful because the
process yields the value of H. Dividing M by a constant, H, does
not change the parabolic characteristics of the function.
H must be found.
value.
First
The maximum height of the arch can be set to any
Let the arch be symmetrical with span, L, and with the
maximum y value occuring at the Y-axis intercept, where x = 0.
Let y = L/2 at this point.
H = Mxy
= PL2
= pL2 /8 when x = 0. Therefore:
Mx
=
pL/4
Eqn. 3-14
L/2
y = Mx/H
= -px 2/2 + pL2 /8
pL/4
y = -2x2/L + L/2
Eqn. 3-16
With Eqn. 3-16 we can plot the rest of the arch.
x = L/4, y = 3L/8;
Eqn. 3-15
For example at
at x = 3L/8, y = 7L/32 (Fig. 3-3).
As the sides of the parabolic arch approach the supports, the
70
PX =
LLL
LL
p
[LLLI
LL L
LL
Y-axis
AH
B
A
i
(0, 0)
B
H
I
X-axis
y = -2X 2/L
Fig. 3-3
+
L/2
Diagram of distributed dead load for a parabolic arch.
curvature diminishes and the arch straightens out.
As the force
resultant due to vertical loads from above gets larger towards the
supports, the addition of the same uniform load continues to change
its direction downward, but with less and less degree of change.
The
arch which allows this force resultant to lie near the central axis,
maintaining the arch in compression, is that curve whose legs
straighten out with the greater distance from the apex (Fig. 3-4).
The parabolic arch is right at home under such uniform loadings
71
uniform load
= p/unit length of span
central
axis
R2
CK1
R1
R2
OC2
P
R3R2
R33
0C 2
Fig. 3-4a)
Diagram showing the relationship of uniform load
and force resultants for the parabolic arch.
Force vector diagrams of two different points
on the arch.
b)
where:
note:
2
p
=
uniform load
R
=
force resultant representing the addition of
previous force resultants and load
CK 2 <
1
holds true because the curvature diminishes
towards the supports
as bridge decks and flat floors and roofs.
The catenary makes good
structural sense in applications such as roofs for large auditoriums
or sports stadiums where the weight of the roof is the major load.
Speaking purely in terms of structural efficiency, the circle is
appropriate where the dead loads approximate the loading shown in
72
Fig. 2-5.
The surmounting layer of earth over an underground drain
pipe is near typical of such a loading.
This is true partly because
of the-orientation of the vertical load and partly because the earth
supplies a force component perpendicular to the surface of the drain
pipe.
For the circle, pure compression or pure tension result from
uniform loadings placed perpendicular to the surface (Fig.
1-6).
Because of the relative ease of fabricating structural circles, the
curve has applications in many instances where the dead load is less
favorable.
As with all curves, aesthetic reasons may supersede
structural rationale, justifying selection.
Superimposing the
semicircle, semiellipse, catenary, and the parabola and their most
appropriate loadings presents a relative picture of the interrelationships.
The circle is the only one of these curves which is somewhat
inflexible.
All the rest have the ability to accept any maximum
height, h, for any given span, L. Due to the spatial limitations of
many architectural programs, this characteristic places a slight
handicap against using the circle.
If the program height for a
space is less than half the span, segments of the circle or sphere
are usable.
But, if the height is greater than L/2, the difference
in elevation can be made up only with a supporting drum.
In the
diagram below I have allowed the height to be greater than L/2 to
enable the semicircle (raised on a drum) and the semiellipse to
participate in the comparison (Fig. 3-5a).
The span and the
enclosed area are consistent for all four examples.
73
Fig. 3-5b shows
loading for
semiellipse
loading
for semicircle
loading
for catenary
loading for
LLI 1 1
/
parabola
1
L VV
El
V
parabolic arch
- catenary arch
-
semielliptical arch
semicircular arch
Y-axis
B
AH
(0, 0)
B
H
X-axis
BV
lA y
Fig.
3-5a
Superimposition of arch forms and the most efficient
relative loadings.
All arches have similar span
and enclosed area.
74
Y-axis
semielliptical
arch
A(00)
catenary
arch
segment of
circle arch
parabolic
arch
BH
X-axis
lAy
Fig. 3-5b
Byr
Superimposition of arch forms. All arches have
the same span and maximum height.
the four curves as arches where the parameters of span and maximum
height are constant for all.
The easiest of the loads in Fig. 3-5 to conceptualize and work
with is the uniform load.
Because of this reason I would like to
use the parabola to demonstrate a design process for a parabolic
barrel vault with stiffening ribs.
In this problem the challenge
of designing the roof structure of a sports arena is presented.
The
roof consists of reinforced concrete arches spanning 260 feet,
spaced 30 feet apart, which carry a thin reinforced concrete shell
roof.
Ideally, the shape this loading suggests is the catenary.
However, the design process is basically the s-ame for the parabola.
75
The greater facility in working with the parabolic shape and the
uniform load make these the best choice for the sample design
problem.
The layout is presented below in section with the end
buttresses set at different levels (Fig. 3-6).
A clear height of
60 feet exists at the center of the span.
Fig. 3-6
Parabolic roof vault for sports stadium.
The dead load requirement for each arch is a uniform horizontally
distributed load of 3.0 kips/linear foot (Fig. 3-7).
one thousand pounds.
One kip equals
The dead load figure of 100 lbs./square ft.
accounts for the weight of the arch and the concrete shell (where
76
uniform dead
load
3.0 kips/linear
=
pg=
foot
p
60'
485 k
485k
605.2
Xa
120.7'-
=
640.2k
362k
418k
260'
Fig. 3-7
Parabolic arch in section with the dead load shown above
and the resulting total reaction forces indicated below
(horizontal and vertical components shown).
where:
M
concrete
=
AH
AV
=
pxa
B
=
pXb
=
BH
= 150 lbs./cubic ft.), the roofing material, and any
mechanical systems hung below.
The live load is calculated
asymmetrically to design for the "worse situation" loading.
parabolic arch carries the dead load in compression along the
77
The
longitudinal axis, but must support asymmetrical live loads partially
by its resistance to bending.
If there was a live load covering the entire roof, the combined
effect of the dead and live loads would be 4.5 kips/linear ft..
figure allows 50 lbs./s.f. for snow and wind loading.
This
The diagram
below shows the resulting reaction forces (Fig. 3-8).
uniform live load
uniform
dead
6 L
load
1.5 kips/linear
=
=
ft.
3.0 kips/linear ft.
L
L,~L L
Ld
72
7Zk
543k
907.8k
627k
960.3k
Fig. 3-8
Parabolic arch in section with the uniform dead and
live loads shown above and the resulting reaction
forces shown below.
The greatest compression stress would occur in the right leg
of the arch where the maximum force due to loading equals 960.3 kips:
78
compressive stress
= Cr
= 960.3/bd
Eqn. 3-17
b = width of arch member
where:
d = depth of arch member
With an ultimate stress of concrete, f
=
4,000 p.s.i., and a
factor of safety, F. of S. = 3, the allowable stress,
Fc
= 4,000/3
= 1,333 p.s.i.
= 1.333 k.s.i..
With the above
information the cross sectional area can be computed, Axs
A
where:
=
P/Fc
= 960.3/1.333
=
720 s.i.
=
bd:
Eqn. 3-18
P = axial load
The stress due to compression along the longitudinal axis
presents no major structural difficulty. A cross section for the
arch of 18" x 40" would suffice.
A far more serious factor in
structural design is the bending moment due to an asymmetrical load
and its potential to buckle the arch.
Buckling is a phenomenon
demonstrated by standing a postcard on its edge.
Applying a load
with your finger to the edge of the postcard bows the card to the
side.
This is buckling.
when the card collapses.
The critical buckling load is reached
Imagine the moment caused by a live load
occuring totally on one side of the parabolic arch.
effect of snow and wind might produce such a load.
The combined
The reaction
forces resulting from just the live load are shown below in Fig. 3-9.
The dotted line in Fig. 3-9 represents the pressure line of the
live load.
The pressure line indicates the arch whose shape would
resist the trajectory of resultant load forces in compression only.
Inverted, the dotted line represents the funicular shape that a cable
79
uniform live load = 1.5 kips/lin. ft.
LI
L
L.
L
eccentrically placed
ev= vertical
C
eccentricity
ev
r
Sppressurees
1
ine
A
B
17 3.8k
X
13 2.5 k
q
48.5k
spans.
M
=
.6k
1 2.6k
Therefore, the moment will be greatest at these points:
He,
where H
eccentricity.
=
horizontal thrust, and ea
=
vertical
Assume the pressure line passes through the apex.
Consequently, there is no moment at this point and it can be modeled
as a hinge.
If the support connections are also treated as hinges
the model takes the form of a three-hinged arch.
the process for finding the support reactions.
This simplifies
Taking moments about
points A and C in Fig. 3-9 will give two equations with two.unknowns.
Solving these simultaneously will yield the answers:
80
+33M
AH=
+3TMC =
B
-60.4(181K) - 15BH + 260B
-60BH
= 48.5K,
+ 139.3By
AV
= 0
Eqn, 3-20
0
r-
= 132.5K,
Eqn. 3-19
BH
=
AH
112.6K
Reactions in line with the pressure line at the supports:
left support reaction
right support reaction
e
=
=
173.8K
122.6K
at the 1/4 point of the right leg:
M,=
He
v
=
112.6K(15')
=
45 - 30
1690K'
=
15'
Eqn. 3-21
The arch must stand up to a 1690K' moment at the quarter span
on either side.
at all points:
The arch must take the bending in either direction
downward and upward, or inward and outward.
The
arch can be designed so that the cross sectional area varies directly
with the changing moment.
constant throughout:
cross section).
cr
The resulting maximum stress would be
= M/S (where S = bd2 /6 for a rectangular
If the arch is designed with a rectangular cross
section, and the width, b, is held constant, the depth, d, varies
directly with the square root of the moment, M (Fig. 3-10).
The arch could be designed with a uniform cross section, but
this would be extremely wasteful of material.
Here, I will merely
figure out what the cross section of the reinforced concrete arch
would have to be at the quarter span to handle the stress due to
moment.
I will assume a rectangular cross section in making the
following calculations.
The arch must resist bending, and have the ability to withstand
81
uniform live load = 1.5 kips/lin. ft.
L
L
L
L
,
C
L
moment diagram for
eccentric loading
max.
positive
moment
max. negative
moment = 1690k'
arch section conforms
to changing moment
Fig. 3-10
Moment diagram for eccentrically placed live load and
the parabolic arch section that would normalize the
maximum stress resulting from this moment.
buckling as well.
Critical buckling is when a member continues to
deflect with no further resistance under the pressure of a given
load.
The result is collapse.
is the critical buckling load.
The load that induces this effect
The formulas that determine the
arch's capability to withstand buckling are those which govern the
behavior of columns.
The arch can be split at the apex.
can be analyzed independently as a column.
Each side
Since the right side of
the sample arch is longer and more inclined to buckle, I will analyze
82
its behavior and allow the design of its cross section to control
the design of the left side.
The approximate length of the right
leg is 150 feet,
There exist two formulas developed by Bernouli Euler to guide
the design of the arch.
The first describes the critical buckling
load for a column, Pcr, and the relationship it has to the modulus
of elasticity, E, the moment of inertia, I, and the length of the
specimen, L:
P
=
T 2 EI
(kL)
where:
Eqn. 3-22
2
k = the effective length factor reflecting the nature
of the end conditions
I assume k = 1, a value which models the condition of a pin
connection at both ends, rotation free, translation fixed. A pin
at the buttress end eliminates a moment connection and allows the
arch member to be attenuated as it approaches the ground.
room for circulation is thereby permitted.
The top end of the arch
member is attached to theother leg of the arch.
I assume the existence of a hinge at the apex.
fore a three-hinged arch.
Greater
For easier analysis
The model is there-
In reality the top end can translate,
but again, to simplify the design process I assume that it is
translation fixed.
The critical buckling load, Pcr, varies inversely as the square
of the length, L. This interaction is expressed by the graph on the
following page (Fig. 3-11).
Note that as L diminishes there comes
83
P
theoretical
buckling curve
region of
buckling
Load P
slenderness
ratio, kL/r
Fig. 3-11
The buckling curve, the relationship of load, p, to
the slenderness ratio, kL/r.
where;
P
ultimate compressive capacity of the material
=
kL/r
=
slenderness ratio
k
=
effective length factor reflecting end conditions
L
=
length of the specimen
r
=
(I/A) /2,
I
=
moment of inertia
A
=
cross sectional area
the radius of gyration
a point wherE the crushing strength (compressive capacity) of the
specimen cont rols the value of Pcr.
value of P
Obviously, Pcr can't exceed the
the crushing strength of the specimen.
84
The experimental
buckling curve presents lower values of Pcr for any given length, L,
Imperfections in the material of the specimen and distortions in the
actual shape of the specimen account for the discrepancy.
If a column stands perfectly vertical and is loaded smack- in
the middle, above the central axis, then Eqn. 3-22 states that there
exists a theoretical load, Pcr, which will fail the column.
The
load that will fail the column is sufficiently reduced if there is
any initial imperfections in the column such as crookedness.
Crookedness introduces additional stresses caused by the moment equal
to the load, P, times the eccentricity, e:
Pe.
The graph below
indicates the effect of the additional stress (Fig. 3-12).
theoretical
Pcr
(theoretical)
Pcr
(experimental)
--
axial load
P
buckling curve
--
/
p.-experimental buckling
curve for specimen with
I
initial crookedness
I
j
. lateral
Fig. 3-12
initial
eccentricity
(e)
lateral
displacement
(J)
displacement, <(
Comp arison of theoretical and experimental buckling
curv es.
85
PI
Because of the disastrous effect that moment can have on a
column's performance, columns must also sustain moment stresses.
The other relevant Euler equation defines the interaction between
axial capacity and moment capacity of a column specimen:
+
P
P01
where:
1
(M )
- P ()
cr
5 1
Eqn. 3-23
P = axial load (kips)
P
Pcr
= theoretical ultimate compressive capacity of the
material (kips)
=
theoretical critical buckling load (kips)
M = maximum moment (kip-inches)
M
= theoretical ultimate moment capacity of the
specimen (kip-inches)
If we are working with loads that stress the specimen to its limit
(where the left side of Eqn. 3-23 equals 1) then the following two
statements must be true:
a) If the maximum moment increases then the axial load
must decrease to maintain equilibrium.
b) If the axial load increases then the maximum moment
must decrease to maintain equilibrium.
The maximum moment is the moment created by the maximum initial
eccentricity, e. The term,
1
1 - P/P
,
is an adjustment factor
which accounts for the displacements (the additional eccentricities)
which occur under load.
Consider the right leg of the parabolic arch as a column
member.
Given the large initial moment, due to the large eccentricity
86
caused by its curvature, if we design the arch so that it satisfies
Eqn. 3-23, which accounts for moment effects, it should easily
comply with Eqn. 3-22, where the critical buckling load, Pcr, is
established without consideration of the moment effects.
Readily employable values need to be written for all of the
variables in Eqn. 3-23 in terms of the rectangular cross section of
the arch with dimensions b x d, where:
b = width, d = depth.
Since the concrete shell roof will laterally restrain the arch,
buckling will only be possible in one dimension, either toward the
outside or inside of the arch.
The following parameters of steel and
concrete will be used in the calculations:
fy
= 60 k.s.i.
Fst
= 33 k.s.i.
Est
=
fc'
FC
30,00 0 k.s.i.
Ec
=
4 k.s.i.
1.33 k.s.i.
=3,000 k.s.i.
Ast/A c =3%
where:
f
y
=
allowable stress of steel (factor of safety
Fst
=
Est
= modulus of elasticity of steel (kips/s. i.)
fcC
= ultimate stress of concrete (kips/s.i.)
Fc
= allowable stress of concrete (factor of safety
E
= modulus of elasticity of concrete (kips/s.i.)
Ast/AC
note:
yield stress of steel (kips/s.i.)
= ratio of area of steel to area of concrete
Throughout calculations all terms remain in inches.
87
= 1.8)
3.0)
Steel reinforcing will be located in both the top and bottom
layers of the arch to restrain bending in either direction (Fig. 3-13).
b---
-U
c
steel
reinforoing
bar
concrete
central axis
(vertical)
d
- -p
C
Fig. 3-13
Cross section of the parabolic arch member.
where:
c
effective moment arm of steel about the
central axis (c = .4d)
=
The area of steel must be written in terms of an equivalent
area of concrete in order to establish a value for the term
P0 .
Steel will have an effective equivalent area of concrete 8 times
the area of the steel:
8Ast
P
o
=
=
Eqn. 3-24
Ac equivalent
A (1 + 8(.0
cc
3 ))Fc
Eqn. 3-25
88
P0
=
1.24A c(1.33)
=
1.65A c(kips)
Eqn. 3-26
P0 is now written in terms of the allowable load because the factor
of safety has been introduced
The theoretical allowable moment, M , is figured in terms of
the effective moment of inertia of steel, Ist, about the central
axis:
Ist
= 2(.015Ac(.4d) 2 ) =
.0048Acd2
Eqn. 3-27
Eqn. 3-28
M
=
Fst (Ist/.4d)
M
=
33(.0048Acd 2 /.4d)
=
.4Acd (kip-in.)
Eqn. 3-29
To find the critical buckling load, Pcr, the terms on the right
of Eqn. 3-22 must be defined:
Pcr
=
=
EIcomp.
where:
Ec c/5 + Est Ist
= bd3 /12
=
EIcomp.
= Acd 2 /12
Eqn. 3-31
2
3,000(Acd 2 /12) + 30000(.0048Acd )
Eqn. 3-32
5
EIcomp.
=
194Acd2
L = 150'(12in./ft.)
Pcr
Eqn. 3-30
= composite EI factor where the EI of concrete
is diminished because of its weaker strength
despite its larger I
EI comp
Ic
Eqn. 3-22
IT 2EI/L2
-
22T
(194Acd 2)
Eqn. 3-33
= 1800 (in.)
=
.0006Acd 2 (kips)
Eqn. 3-34
Eqn. 3-35
(1800)2
The maximum moment was calculated before in Eqn. 3-21:
M = 1690'(12in./ft.)
= 20280 (kip-in.)
89
Eqn.
3-36
The axial load, P, is equal to the reaction load along the
longitudinal axis due to a uniform dead load of 3.0 kips/ft. plus
the reaction load due to an eccentric uniform live load of
1.5 kips/ft.:
P
640.2 + 122.6
=
Eqn. 3-37
762.8 (kips)
=
With the terms defined, Eqn. 3-23 can be restated:
+
P
1
P0
(M ) !
Eqn. 3-23
1
1- P
(M )
P
cr
762.8
1.65Ac
(20280)
(.4A cd)
c
.0006Acd
1
1 - 762.8
+
462.3
Ac
1
(50700)
1 - 1271333.3 (Acd)
A d2
c
+
!
4
Eqn. 3-38
1
Eqn. 3-39
1
Eqn. 3-23 is now in a form where potential values for b and d can
be plugged in and checked.
To begin,
I assume b = 18", d =
90
With these dimensions, the resulting influence of the stress due to
axial load and the moment stress can be found:
+
462.3
(50700)
1
1
1
Eqn. 3-40
1 - 1271333.3 18(90)2
18(90)
18(90)3
.29
.68
+
!
.39
1
1
Eqn.
1
3-41
Eqn. 3-42
OK
The moment term, .39, is responsible for .57 of the controlling
factor.
Yet, these dimensions are well within the limit.
If b
remains constant and d is diminished, there will be a change in the
relative proportions of influence.
90
The stress due to moment varies
as the square ofthe depth, while the stress due to load along the
longitudinal axis varies directly with the depth.
Assume:
d = 76":
b = 18",
.34
+
.58
.92
a
1
1
1
Eqn. 3-43
OK
Eqn. 3-44
As predicted, the moment term,.58, now contributes to .63
of the controlling factor.
These dimensions are quite safe.
Another 4" reduction in the depth and the -limit would be surpassed.
The additional margin of safety represented here, more than
compensates for the minimal errors in the forming of the parabolic
arches.
With the cross section determined, the area of the reinforcing
steel can be calculated:
Ast
= .03Ac
=
.03(18)(76)
41 s.i.
Eqn. 3-15
Seven #16 reinforcing bars (2" diameter) set in the top layer of
the arch and seven #16 bars set in the bottom layer will more than
supply the necessary area.
Because the moment decreases to zero near the supports, the
depth of the arch can be reduced accordingly.
But. the arch must
maintain sufficient cross sectional area to withstand the maximum
axial load at the buttresses (Fig. 3-8):
Ac + 8A
Ac + 8(44)
Ac
=
P/Fc
=
Eqn. 3-16
960.3/1.33
= 370 s.i.
Eqn. 3-17
Eqn. 3-18
91
If the width, b = 18", then the arch member can be reduced to a
depth, d = 22", at the buttress.
The shell interconnecting the arch can be dimensioned in the
following manner.
Since the thinness of the shell makes it
susceptible to damage, the allowable stress of concrete, Fc, of the
shell itself is set at 1/3rd of the allowable stress used for the
arch:
Fc
Eqn. 3-49
1 (1.33 k.s.i.) = .44 k.s.i.
3
The maximum stress the shell will experience will be at the bottom
Inspecting a one foot wide strip from the apex to the
edge.
bottom edge will provide the information necessary to calculate the
potential maximum force at the bottom edge.
Eqn. 2-11, Hy
=M
,
when manipulated gives the horizontal thrust, H:
H =
Eqn. 3-50
M /y
If the maximum moment and the maximum height of the arch are plugged
in as values for M and y, Eqn. 3-50 can be rewritten:
where:
H
=
pL2 /8h
p
=
dead and live load per horizontal linear foot
of the strip
L
=
span of the strip
h
=
maximum height of the strip
Remember, Eqn. 1-4, states: R
Eqn. 3-51
=
L2 /8h.
Substituting R0 in
Eqn. 3-51 above results in:
H
=
Eqn. 3-52
pR0
The force maximum in the one foot strip is at the bottom edge
92
where the angle of the shell with the horizontal is approximately
Therefore the force, T1 , at this point is found by Eqn. 1-10
450
shown below:
= H/cosoC
T1
and:
F
where:
A
=
=
T/Axs
t
=
=
=
=
pR0/cos 450
=
1.4pR
Eqn. 3-53
Eqn. 3-54
1.4pR0/Axs
area of the cross section of the one foot strip
of shell
=
xs
Since F
H/cos 45
.44 k.s.i. for the shell, and, A
=
12" x t (where
thickness of shell), Eqn. 3-54 becomes:
Eqn. 3-55
.44 = 1.4pR 0/12t
t/R
=
Eqn.
.27p
3-56
The ratio expressed in Eqn. 3-56 is appropriate for any parabolic
barrel vault.
In this example, where the dead and live load equals
150 lbs./ft. (.15 kips/ft.) of the one foot wide strip:
t/R0
and:
R
= .27(.15) = .04
= L2 /8h
t
therefore:
=
= (260)2/8(52.5) = 161'
6.5"
Eqn.
3-57
Eqn.
3-58
Eqn. 3-59
Of course, if the stiffening ribs are any closer than the 30
feet used in the example then the thickness of the shell can be
reduced.
Where the shell ends 15 feet above the ground, the edge will
have to be thickened into a beam to transfer the load forces to the
arch members.
The depth of the beam should equal half of its span:
d = span/2
= 30/2
= 15'
93
Eqn. 3-60
Reinforcing bars appropriate for the loads and the resulting moments
The final arch
are a necessary part of the strength of these beams.
and shell sections, based on these design decisions, are shown
below (Fig. 3-14).
t
= 6.5"
15'
t
15'
16
15 '
section
through
shell
-
260'
section
Fig. 3-14
through arch
Shell and arch sections with dimensions.
94
With this brief glimpse at the mathematical definiti-ons of
several curves and the resulting structural logic behind the various
loadings associated with each curve, a better understanding of the
architectural merit of their applications is available. In the next
chapter I will investigate several kinds of vaults.
I will focus
on smaller spans, within the ten to forty foot range.
These
dimensions encompass the range where the cost of building the formwork often precludes the curved ceiling as a possibility.
The
exploration will attempt to pan out the more economic solutions of
the methods which exist.
The emphasis on vaults grows from an
acceptance of the inherent beauty of vaulted spaces.
The methods
of construction must lend themselves to exposing the underside of
the vault to view
with
the potential of employing the structural
strength of the arch form to support a floor or roof above.
Consideration of the rectangular floor plan and its effect on the
shape of the vault will largely influence this search.
It is from
this conventional and spatially efficient container that many of
these vaults spring.
95
96
IV.Vaulted
Ceilings
In this section I will investigate the constraints that program,
in the spatial sense, and construction methods place on the forms
with which one chooses to design.
I will introduce the study with
a reflection on the philosophy of aesthetics and what psychological
reasons it suggests for the selection of particular forms.
I will
then probe the viability of several kinds of curves in vaulting used
to span various rectilinear room plans.
plan for numerous reasons.
used in room design.
I focus on the rectangular
First, it is the most common geometry
Second, the efficiency of the rectangle as a
space giving container has been well proven.
It is easily joined
to other similar shapes and provides maximum functional performance
where a high density of building units is common as in urban
environments.
Windows and doors are simply placed in the planar walls
of a rectilinear volume and right angle corners are economical to
build.
The construction industry and interior furnishings manufacturers
are geared for this kind of living container. Thirdly, combining
nonrectilinear
spanning ceilings and roofs to rectangular rooms
presents a significant challenge to the imagination and one's
physical resources.
97
Today, primary reasons for designing vaulted ceilings where the
spans are less than 40 feet are aesthetic.
This is true because over
this distance modern post and beam framing is the cheapest solution.
In past history what have been some of the intellectual reasons for
building with or without vaulted ceilings?
The Romans had their aesthetic and structural reasons for
consistently building with a semicircular arch.
As already mentioned,
structurally the semicircle was quite successful within the application
of the aqueduct, where the loading was fairly appropriate.
This
fact, along with the ease of cutting voussoirs each with the same
taper, influenced its usage for other constructions.
the circle represents harmony and unity of form.
Aesthetically,
With a semicircular
arch the line of the curve flows smoothly into the vertical line of
the supporting column or pier.
In the Medieval Age when stone was the medium for building
churches, the vault was the only structural form which could
conceivably support stone over spans up to 40 feet.
There certainly
were religious overtones in the choice of the lancet arch for Gothic
churches.
But, the pointed crown may have also had structural
justification.
A lancet arch is a logical form to support a large
load at the crown, such as that of a surmounting wooden truss used
to hold up the roof (Fig. 4-1).
There existed more primitive reasons both before and after Christ
as to why man has chosen to shelter himself under vaulted forms.
Man for thousands of years was a troglodyte.
98
Psychologically, modern
Fig. 4-1
Masonry lancet arch supporting wooden roof truss in
Gothic church.
man still is a troglodyte.
The caves of today are simply dusted a
bit more often than in times past.
. . . the human skull.
Man's mind holds court in a cave
It seems quite natural that we would feel
at home in an environment which is physically reflective of our mind
space.
There are several Olmec sarcophagi in the Museum Parque La Venta
in Villa Hermosa, Mexico which capture the poise of body and mind
which the cave gave man in this epoch (800 - 400 B. C.).
The
sarcophagus illustrated below is a striking representation of the
inter-workings of man and his world (Fig. 4-2).
The noble lord picture in Fig. 4-2 sits alert, yet restful in
a cave hollowed out of a massive chunk of basalt rock, his world.
99
Fig. 4-2
Olmec sarcophagus from La Venta, Mexico, 800 - 400 B. C.
(now resting in Parque La Venta, Villa Hermosa, Mexico).
He supports the world with its surmounting slab to the same degree
that it sustains and protects his presence.
The above slab carries
the incised lines of a jaguar's face suggestive of the underworld
or world of the unknown.
qualities.
For the Olmecs, the jaguar held mythical
Mutants born into their society were believed to be
the product of a union between a woman and a jaguar, and were elevated
to the status of priests.
The rock dwarfs the man.
And yet, he
grasps a thick rope which encompasses the entire rock showing his
100
control over his environment.
His right hand clasps the rope while
his left hand grips his own ankle suggesting the connectedness of
his being to his world.
portrayal.
The work is a masterpiece of character
The engagement of the cave is what makes it sing.
The
little hollow denies the solidity of the rock allowing the noble
a protective place within while still maintaining his controlling
presence without.
A primitive form albeit a sophisticated image.
The Mimbre Indians of the Southwest (region of present-day
eastern Arizona and western New Mexico, 300 - 700 A. D.) closely
associated the curved forms of their pottery as representations of
their mind and soul spirit.
They are well known for piercing the
center of a dead man's pots and placing these along with the dead
man in his grave.
The ritual act was supposed to aid the soul's
passing from the body.
The Greeks and the Maya both knew about curved forms for both
civilizations used curves in plan, but preferred other forms for the
elevations and vertical sections of their temples.
The Greeks
employed a post and beam arrangement which greatly limited the
spacing of columns.
Witness, the Parthenon.
The Maya built a
circular (in plan) observatory at Chichen Itza Viejo (classic
period, 300 - 900 A. D.), but the vertical sections of their temples
were almost always constructed in a truncated A-shaped stone arch
(Fig. 4.3 and Fig. 4-4).
The subject of the Maya (700 B. C. - 1400 A. D.) is a fascinating
one for it embraces the whole dialectic of aesthetic of form versus
101
structural logic.
Here was a people who had a fully developed
written language, an understanding of the stars, an accurate calender,
and a sophistication in the science of building that is uncontested.
They could have selected nearly any form they desired, structurally
or unstructurally sound, and figured out a way to build it. Turns
out, they chose an A-structural form, one that has some structural
logic, although not a real winner given the fact that they built
almost entirely in stone.
The span of the A-formed arches was usually very small, generally
less than 10 feet.
The shape was not excessively strong considering
the large distributed loading it was placed under and given that the
pressure line for the load is something akin to a parabola.
Many
historians have stated that the Maya were not aware of the potential
of the voussoir arch.
This author is of the opinion that the Maya
knew of curved arches, but that the nature of their spiritual
allegiance led them to build the A-shaped arch.
In fact, I have found
in the ruins of Chichen Itza Viejo ("Old Chichen," classic period,
300 - 900 A. D., before Toltec influence) a voussoir arch supporting
a passageway of a building near to the Observatory.
It is unlikely
that this was an archaeologists reconstruction for all of the other
reconstructions have been made using variations of the corbel arch.
In addition, as mentioned, the Caracol or Observatory was built in
the round.
Therefore, the Maya were obviously capable of conceptual-
izing in curved shapes.
Because the Maya's choice for an arch form was not ideal for
102
Fig. 4-3
Maya stone corbel arch, Palenque, Mexico, classic period
300 - 900 A. D..
the loads they developed techniques of construction, some rather
ingenious, to keep them from tumbling down.
Many of these arches
are of the simple corbel type where each stone slab cantilevers
beyond the one below it and is counterbalanced by the load above
(Fig. 4-3).
A more finished version and one which was more cleverly
engineered is witnessed below in Fig. 4-4.
Here the finished stones
of the arch are basically isoceles triangles shaped with the long
103
Fig. 4-4
Modified Maya stone corbel arch, Chiche'n Itza Viejo,
Mexico, classic period 300 - 900 A. D..
side of the triangle carved concavely.
The vertex between the two
similar sides of the triangle rests on the stone below with one leg
providing the face of the arch while the other is imbedded in the
loose rubble fill.
lever.
The leg which juts into the rubble acts like a
As the loose rubble settles its weight presses down on this
lever which causes the end of the opposite leg to press up on the
stone above.
The upward pressure resists the tendency of the flat
104
face of the arch to collapse inwards.
The aesthetic choice of the
Maya may have been justified by their religious beliefs, but it
certainly presented their masons with a challenging problem. And
physically, the choice of a truncated A-form arch severely limited
the width of their rooms and passageways.
Throughout history there were many reasons for building vaults
and arches, and many of these had no structural rationale.
twentieth century is no different.
The
In the two photos below a
structurally efficient vault is juxtaposed to a structurally inefficient
arch (Figs. 4-5a and b).
Fig.
4-5a
Pension with repetitive barrel vaults for a ceiling,
Oaxaca, Mexico, built 1970.
105
Fig. 4-5b
Sewing Factory, Paris, France, by Auguste Perret,
completed 1919.
The repetitive barrel vault of a small pension in Oaxaca, Mexico,
built in 1970, is employed as a cheap method to span from wall to
wall of each room (Fig. 4-5a).
It is the appropriate form for the
loading and consequently allows the tile and concrete ceiling to be
quite thin.
In the other example, Perret's sewing factory in Paris,
completed 1919, the concrete arches are not the appropriate funicular
form for the load (Fig.
4-5b).
As a result the arches must absorb
106
considerable bending force.
Who is to say that the expense of both
solutions wasn't justified?
The pleasure that Perret's curved arches
have given the workers of the factory is immeasurable.
Perhaps
productivity was improved because of this elegant solution.
What would I as a present-day designer put forth as a rationale
to justify the use of vaulting in office or residential construction?
The principle reason I have is that vaults are less monotonous than
flat ceilings.
They afford a dynamic change at or above the molding
line where the ceilingmeets the wall.
The
chiaroscuro of a curved
surface is more subtle than that of a flat surface. There is more
gradation of light intensity, a greater diffusion of shadow.
Vaulting
tends to deliver a hierarchy to the various positions within the
enclosed space.
Notions of axis, dominance - subdominance, and
centrality of space arise in association to vaulted spaces.
These
along with the structural aspects of vaults are some of the reasons
why they have been built so often in churches.
I believe there exists
a fundamental concept which underlies the above mentioned feelings
about vaulted ceilings.
Man is essentially at'home in nature.
Nature in its wondrous
complexity presents us with objects and creatures which exist in
innumerable shapes, sizes, and textures.
The diversity of subject
and form found in the natural world are an inspiration and delight
to mankind.
in nature.
There is balance in this diversity.
Curves are plentiful
Imagine a tree-lined street in your home town where the
arching branches form a bower overhead.
107
Curves are much less
apparent in the modern urban built world of glass, steel and concrete.
Not only is there a loss of physical touch with the earth and its
fruit in these environments, but there is also a loss of intellectual
stimulation by the lack of many of the common earth forms.
If one
were to introduce more curves into the ubiquitous Miesian cubicles
which abound in today's designs and yesterday's buildings
.
where
.
to logically place them?
Warping the floor might provide good exercise, but would be
unfair to octogenarians.
And if the warp presented a convex ceiling
to the inhabitants below the result could be untenable on psychological
grounds.
Nature does have its share of unpleasant curves.
the walls in plan or section might be nice.
Curving
It also might be quite
impractical spatially for the user, the surrounding rooms, and the
interior furnishings.
Arched windows and doorways and curved relief
work in planar walls is a different issue.
ceiling.
Finally, voila the
The presence of the ceiling is mostly a visual one.
Rarely
do we touch the ceiling or bump up against it. What better place
for a nonlinear surface?
It is possible to construct vaulted ceilings without raising
the ceiling height.
The residual spaces between vault and floor
can be used to house mechanical equipment.
There is a definite
practical potential which goes hand in hand with these less than
flamboyant forms.
1)
Two questions will shape the remaining discussion:
Volumetrically, for the rectangular room, what are
some of the vault configurations available to the
designer?
108
2)
Given the difficulty of building certain forms,
what vaults are best built as false ceilings and
which can logistically take on the structural role
of supporting themselves as well as a surmounting
floor or roof?
A look at smooth volumetric contours will open up this survey.
Later, ribbed vaulting and its variations will be introduced.
The
simplest of vaults is probably the barrel vault with single curvature.
In one direction the section is curved while in the perpendicular
direction the section is rectilinear (Fig. 4-6).
The barrel vault
can have any kind of curvature as long as the chosen curvature is
uniform for the length of the room.
Barrel vaults can work as beams to span from wall to wall or
column to column.
They have the ability to act like a beam when
spanning along the length of the barrel vault.
If a single barrel
vault is employed to span a room a diaphragm or tie member is necessary
to restrain the horizontal thrust of the cross-span arch of the
vault.
If barrel vaults are set side by side the horizontal thrust
of a vault can be counterbalanced by that of an adjacent vault
(Fig. 4-7).
Of course at the end bay a buttress or tie member
through to the opposite side is necessary to restrain the thrust.
If two similar barrel vaults intersect at perpendiculars at the
same altitude we have the vault geometry that is shown in Fig. 4,8.
The illustration depicts the vault set over a square room plan.
Extrapolating this geometry to a rectangular room plan produces the
image presented in Fig. 4-9.
Any one of these curved surfaces can
be inclined with regards to the horizontal to arrive at variations
109
B
A
A
r-B
Ir
1:
Section
Fig. 4-6
A-A
Section B-B
Axonometric and sections of a barrel vault set over
a rectangular room plan,
110
/
Fig. 4-7
Barrel vaults used as spanning elements like beams.
Repeating the vault side by side enables the horizontal
thrust of a vault to be counterbalanced by its neighbor.
111
Fig. 4-8
Two barrel vaults set perpendicular to each other
intersecting at the same altitude.
to the configurations shown in Fig. 4-8 and Fig. 4-9.
If the curved surfaces which intersect are not sections of
barrels but sections of saddles (negative total curvature - doubly
curved in opposite directions) the vault attains the character of
112
Fig. 4-9
Variation of the vault geometry depicted in Fig. 4-8, here
designed to surmount a rectangular room plan.
113
the outline drawing reproduced in Fig. 4-10.
The figure is actually
two hyperbolic paraboloids which intersect at ridges which diagonally
connect the four corners,
An interior perspective displaying this
geometry as a vaulted ceiling over a dance studio is portrayed in
Fig. 4-11.
Because of the double curvature the last illustration
is a more difficult one to build than the previous examples.
M~i
Fig. 141
Fig.
4-10
Two hyperbolic paraboloids intersecting at ridges which
diagonally connect the four corners.
114
Fig,
4-11
Interior perspective illustrated with the vault geometry
depicted schematically in Fig. 4-10.
115
A simpler doubly curved surface with negative total curvature
is the conoid.
Professor Waclaw Zalewski defines the conoid with
the following words and illustration (Fig. 4-12):
A conoid results when a straight line glides along two
guiding lines one of which is curved and the other straight.
Conoids belong to the family of membranes with a negative
total curvature, which can be detected by observing the
intersection of the conoid surface with a slanted plane. 4
Fig.
4-12
Conoid intersected by a slanted plane to show
negative total curvature.
The conoid vault, designed for use in an office organization,
is displayed in Fig. 4-13.
The height of the vault can be adjusted
to conform to program restrictions while still allowing sufficient
Zalewski, Waclaw P. Membranes and Shells p.
116
37
dimension to the residual space so that it can house portions of
the mechanical system,
Fig 4-13
Office organization surmounted by conoidal vaults.
Another possibility for a rectangular room is the hyperbolic
paraboloid (hypar), either one or combinations of more than one.
A ceiling/roof made from four hypars is drawn below in Fig. 4-14.
Next is a vault which has positive double curvature where both
perpendicular sections have curvature in the same direction.
One
can conceptualize a variation of this form by imagining a thin
rubber membrane stretched over a rectangular frame and inflated.
Many tennis courts with inflated membrane roofs have this configuration.
If the curvature in both directions is noncircular, the
117
solid diagonal lines
dotted
diagonal
Fig. 4-14
lines
principle
=
=
principle
lines of
lines of
compression
tension
Ceiling/roof made from four identical hypars set
against one another. All edges between the hypars
as well as the outer edges are in compression.
curves of either direction will have changing curvature as the
sections are cut from the middle of the volume toward either edge
(Fig. 4-15).
As depicted in Fig. 4.15, the curvature of the middle sections
118
ourvature
of section
K0
>
K1
K2 > K
3
Fig. 4-15
Vault with positive total curvature surmounting a
rectangular room. Curvature in either direction is
noncircular -(perhaps of a parabola or a catenary).
Greatest curvature lies with sections Ko amd K 2 in
the middle of either side. Least curvature lies
with sections K and K3 toward the ends of either
side.
is greater than that of the sections toward the ends of either side.
In other words, the radius of curvature, Ri (R. = 1/K.), of the
119
middle sections is smaller than that of the sections toward the end
of either side.
A variation of the vault shown in Fig. 4-15 is a vault which
has curvature in both directions, but where the curvature is
constant across both the width and length of the rectangular plan
(Fig.
4-16a and b).
curvature
Fig, 4.16a
of section
KO = K
1
Vault over a square plan in which each direction has
constant curvature over its width, K
=
K . Ridges
are a necessary facet of this vault.
120
curvature
KO = K1
K2 = K
3
Fig. 4,16b
Vault over rectangular plan in which each direction
has constant curvature over its width and length,
K
= K
K
= K .
3
2
1,
o
121
In both Fig. 4-16a and b ridges form where the two contours
intersect.
If the vault is to have stiffening ribs these ridges
provide logical lines along which to place a set of diagonal ribs.
If the room is rectangular the vault takes the form depicted in
Fig. 4-16b.
Additional ribs can be aligned orthogonally to the
room plan at regular intervals inbetween the diagonals.
It should
be clear that the vaults pictured in Fig. 4-16 are much easier to
construct than the vault shown in Fig. 4-15 because of the consistency
of the curvature across the width and length of the vaults in Fig.
4-16.
Moving again to more complex vaults it becomes quickly apparent
that there exists a great number of possibilities for developing
complex vault shapes.
I will refer to a few of the more notable
remaining geometries in the following illustrations.
A ribbed vault which was employed commonly in Gothic ceiling
design is pictured and described below by Walter C. Leedy Jr. in his
article, Fan Vaulting
(Fig. 4-17):
Ribbed vault, the classic Gothic ceiling design, was
conceived by medieval masons as a frame structure
in which pointed arches serve as weight-bearing ribs.
Each of the modular units of the vaulting, called a bay,
includes six arches arranged as a rectangle with
intersecting diagonals. The load of the ceiling is
converted into outward and downward thrusts that are
conveyed in part through the arches to the piers and
walls. The surface of the vault between the arches is
filled with stone sheets called webs. In England the
stones were laid at an angle of 45 degrees from
the longitudinal or transverse ridge to the groin.
5 Leedy Jr. Walter C. Fan Vaulting Scientific American Feb. 1973 p. 136
122
Fig. 4-17
"Ribbed vault, the classic Gothic ceiling design."
One can imagine a room accepting the vaulting that
typifies a "bay" in the above example.
It would certainly be an
intense ceiling structure to have over one's office or living room
unless flattened out to a much greater degree than typified in
Fig. 4-17.
A more smoothly contoured vault which can be even more fanciful
than the Gothic vault is a vault which employs four surfaces of
rotation (fan-vault conoids) to a bay or room (Fig. 1).
123
Picture
a curved line which extends between a point and a circle.
If the
line is rotated about the circle it defines a surface which is
doubly curved like a saddle.
Within a bay or room the rotated
surfaces are set on their pointed ends one to each springing or
corner.
The void between their lines of intersection is filled by
a flat horizontal surface (Fig. 4-18).
Fig. 4-18
Fan vault created from "fan-vault" conoids.
124
Finally, there are many types of vaulted ceilings which can be
made from linear elements, which when combined form arched surfaces.
In the example illustrated in Fig. 4-19, designed by Professor
Waclaw Zalewski for industrial application in Poland, the vaulting
springs from structural column capitals.
I I 177777=
lot
Ila 11TTT"r=,,,
71,
plan schematic
section,
structural
Fig.
4-19a
diagram
of
forces
components
Multi-story project for industrial use, Poland 1958,
Waclaw Zalewski, engineer.
125
Fig.
4-19b
Axonometric of multi-story project for industrial use,
Poland 1958, Waclaw Zalewski, engineer,
126
Each capital has four structural ribs oriented on the diagonal to
support ribbed concave panels.
from the underneath.
The panels are concave when viewed
The main span of the diagonal ribs is interrupted
by flat horizontal panels set in the middle of a larger arched square
panel,
The remaining voids between adjacent square panels are filled
with hexagonally shaped ribbed panels.
All elements are precast.
After being set in place the seams between components are grouted
to lock the assembly together insuring a continuity of structure.
The horizontal thrust of each bay is counterbalanced by the thrust
of adjacent bays.
To take care of the end bays a tensile tie member
can be placed from one side of the building to the opposite side.
Now, how to build these contoured surfaces?
materials?
And with what
To build economically it would be necessary to place a
few restrictions on the material and the method.
I don't believe one
can build inexpensive vaults today in stone, brick, or tile.
Their
weight would demand considerable buttressing to counteract the
horizontal thrusts.
But more important perhaps as a construction
expense would be the necessity to use falsework while setting the
pieces of the vault.
The number of man hours required to rig up the
falsework and set the pieces of the vault in mortar can be enormous.
Concrete too, necessitates falsework.
advantages over stone and brick.
But, concrete has two
First, concrete mortar can be
pumped to difficult locations through hoses by a pump truck.
It can
be sprayed onto surfaces (Gunite) and allowed to set up in thin
layers.
The early layers once set give structural support to later
127
layers.
Because the first layer can be quite thin and light it is
possible that the formwork supporting it could be an inflatable
membrane.
The membrane can easily be placed for the spraying and
removed afterwards.
can be reused.
If appropriate precautions are taken the membrane
The second advantage concrete has over other masonry
materials is that it can be precast in readily transportable elements,
The benefits this brings to precision forming, on-site assembly, and
finish work are suggested by Zalewski's work depicted in Fig. 4-19.
For most applications the masonry materials including concrete
are probably the only sensible materials to employ both visually
and structurally for vaulted surfaces.
matter.
Stiffening ribs are a different
Here oftentimes steel and wood are justly employed.
The
advantages and disadvantages of the various alternatives have to be
carefully considered.
For example, concrete can with one surface
provide a vault and a structural support for a ceiling or roof above
(Fig. 4-20).
vault.
Stiffeners are simply cast as one piece with the
Of course concrete requires some kind of formwork and many
times the easy solution won't suffice
Formwork can be reused such
as with slip-forms in a barrel vault.
But, concrete vaults no matter
how thin are heavy and create large reaction forces to counteract
at the points of support.
Wood,
if
it
is allowed by the fire code, is relatively light.
Reaction forces are minimal when compared to those caused by masonry
vaults.
Even though the curve of the vault itself may have to be
made from a nonstructural laminate, the hidden framework can do much
128
Fig.
4-20
Concrete barrel vault with stiffeners with or without
underlying tie members. The structure as shown
is ready to accept a floor or roof above.
to support a surmounting floor or roof.
Whatever the curve of the
vault a truss-like framework incorporating the above structure of
the floor or roof can be constructed to take the loads of the vault
129
as well as the surmounting loads.
If properly designed the truss
like the concrete shell ts capable of withstanding moment.
Therefore
the horizontal thrust at the wall supports can be minimalized if
necessary.
The truss can take moment and still remain quite light
while the moment absorbing concrete shell could require extensive
reinforcing of substantial weight.
Wood trusses supporting suspended
curved false ceilings are pictured below in both monumental and
modest applications (Fig. 4-21a and b).
Fig, 4-21a
Maria Steinbach, Germany, monumental wood truss
supporting roof. False ceiling is hung from the truss.
130
L
Fig.
4-21b
L
L
L
Simple wooden truss engaging the floor joists above
as a structural member. False ceiling hung below.
In Fig. 4-21b the knee braces at either side divide the above
floor span into three, reducing the span and as a result, the size
of the floor joists.
The diagonal braces carry much of the load to
the side framing members placing them in bending.
Stud walls are
quite capable of handling a certain amount of force due to moment.
If the loads are too great for the regular stud member, the points
at which the load enters the stud wall can be beefed up with
additional studs.
If the program allows for it, extra space above
the false ceiling can be taken to construct a more rigid moment
absorbing truss.
Clearly, there are an infinite number of material combinations
and construction methods to an infinite number of possible physical
131
vault configurations.
result.
The cost may or may not justify the end
If the nonefficient structural solutions of modern design
all turned out as gracefully as the arches of Perret's sewing
factory, builders could take every weekend off and architects could
plan their vacations in the Mediterranean.
Undoubtedly curved
spanning vaults for small rectangular spaces, 10 - 40 foot range,
are going to be more expensive to design and build than flat
ceilings.
Yet, the nonmonetary benefits can be great.
Given the
space limitations of most modern multi-story projects, vaulted
ceilings with low profiles are certainly the most reasonable choice
if one plans to design with curved forms overhead.
The main point
is that curved ceiling forms are an alternative for rectangular
rooms, an alternative which like much of nature in the modern world,
has long been overlooked.
132
Conclusion
There is
I assert the importance of the conceptual viewpoint.
the world of exact physical distances and precise object sizes.
do not assume that this is the visual world.
a significant role.
I
However, it does play
I believe that what we see largely consists of
what we imagine we see.
The architect's job is extremely difficult
because there exist innumerable people, like myself, each with his
own imagination.
The designer first inspects his own beliefs about
the nature of the project with which he is confronted.
He must then
infuse his conclusions into a design which will transmit his ideas
to the greatest number of people.
The designer of a modern office
building might ask himself what life would be like in an office of
his building? What is essential about this kind of space?
Is the
space to be nonspecific, principly defined by the two slabs and four
walls between which it rests, or will it have a form which
differentiates places within the room?
What will order
the sense
of space within these offices?
By allowing myself a reference to two non-architectural objects,
I would like to infer that his answer might be colored by the images
presented by two Renaissance artists.
133
We have before us, "The Three
Graces," (1638 - 40) by Rubens, and "The Adoration of the Shepherds,"
(1603-7)
by El Greco (Figs. C-1 and C-2).
In the Rubens' painting, the figures cling to one another
engaging each other's flesh and attention.
All three face inward,
left foot forward, right leg bent, with just the ball of the right
The characteristics of position mentioned
foot touching the ground.
The
above signify the attitude the artist toward the composition.
manner by which he conveys the expressive qualities of the configuration is inherent in his interpretation of the depth of the twodimensional canvas.
The figures interlock.
They form a circle.
There is nothing significant about these facts.
Yet, there is a
By
disposition about the central figure particular to this work.
the strength of her stance between the others, she orders the frontal
facings of her two companions.
Nonetheless, her
body floats
both to the front and back of those adjacent to her.
The
phenomenon is partly due to the orientation of the side figures.
Their broadest dimension, shoulder to shoulder, aligns with the deep
space of the background.
The central figure's broadest dimension
is coplanar with the surface of the canvas.
Consequently,
her
body appears flattened and the least defined within the threedimensional representation.
Also, significant to this effect, is
the ambiguity of the background shading about her person.
The right
figure is fixed against the dark foreground of the woods.
The
left figure is set against the light background of the sky.
The
middle figure oscillates between the other two, forward and backward,
134
Fig. C-1
''The Three Graces,' 1638 - 40, Rubens (painting
rests in the Prado Museum, Madrid)
135
IV
Fig. C-2
"The Adoration of the Shepherds," 1603 - 7, El Greco
(painting rests in Prado Museum, Madrid)
136
by virtue of her position along the edge of the shallow and the deep
space.
By contrast, the central figure of El Greco's "The Adoration of
the Shepherds" occupies the primary position, physically and
expansively.
There.is a hollow of space about the nest of the Christ
child which is suggested by the encircling wall of shepherds and
the dome-like cluster of the hovering angels.
The encompassing
figures are separated from each other, segmented like the piers which
mark the side chapels of a small church.
These are the qualities
by which El Greco envisioned the model for his centralized design.
Thet.echnique which transmits his inspiration to the viewer is the
character of the light which emanates from the Christ child, suffusing
the features of the surrounding figures.
Here the effect of the
chiaroscuro is not unlike that achieved in certain Baroque churches,
where the opening to the lantern of the dome creates a light source.
In the Rubens and El Greco paintings there are depicted two
very distinct approaches to centrality.
a woman is the center.
In the first, the body of
Her physical energy engages the arms and
bodies of her companions.
Visually, the ambiguity of depth about
her person suggests the presence of a void.
In the second, the
Christ child is a point source of illumination, radiating light onto
the inner sphere of the encircling forms.
To make generalizations
which broadly categorize the outlook these two artists held toward
composition would discredit the value their work has for the public at
large.
Their worth lies in the specific details of brush stroke
137
and subject portrayal through which the artist entrusted his
contemplations to the common person.
true for architectural works.
I believe the same holds
After my efforts to attend to the
peculiarities inherent in curved spanning structures, I think it
would be imprudent to conclude with sweeping generalities.
I have
attempted to make small departures within each chapter to
summarize my intentions of analysis.
My conclusions about the design
preparations of architects have a greater impact when mentioned in
conjunction with their work.
I believe that curves have a place
in the future work of architects for their spirit is sorely missing
in the work of present-day architects and engineers.
The farther
one goes in breadth and height from the earth the more necessary is
their presence, structurally and visually.
138
139
140
Bibliography
Billington, David
(Princeton, 1979)
Robert Maillart's Bridges "The Art of Engineering"
Princeton University Press
Billington, David
(Princeton, 1978)
Structures and the Urban Environment
Princeton University Press
Boylan, Barbara Jeanne The Odyssey of Intuition "A Non-Reductive
Interpretation of Technology through a Case Study of Bridges"
(Cambridge, 1982)
Francis, A. J. Introducing Structures
Press Inc.
Pergamon
(Oxford, 1980)
Gheorghin, Adrian; Dragomir, Virgil Geometry of Structural Forms
trans. Dinu Mangil (England, 1978) Applied Science Publishers Ltd.
Gudiol, Jose
El Greco 1541 - 1614
(New York, 1973)
The Viking
Press Inc.
Hardoy, Jorge E. Pre-Columbian Cities
(New York, 1973)
Walker
and Company
The Book of Bridges "The History, Technology and
Romance of Bridges and Their Builders" (England, 1976) Marshall
Hayden, Martin
Cavendish Publications
141
Irvine, Max H. Cable Structures
University Press
(Cambridge, 1981)
Ivanhoff, Pierre Monuments of Civilization Maya
Grosset and Dunlap Publishers
Joslin, Alan Royal Inhabiting the Square
and Place" (Cambridge, 1982)
Krishna, Prem Cable-Suspended Roofs
McGraw-Hill Inc.
Leedy Jr., Walter C. Fan Vaulting
1983 Vol. 248 No. 2
M.I.T.
(New York, 1973)
"A Geometry for Path
(New York, 1978)
Scientific American
Les Albums D'Art Druet XVI A. & G. Perret
de France
February,
(Paris, ?) Libraire
Podolny, Teng Design Fundamentals of Cable Roof Structures
(Pittsburgh, 1969) U. S. Steel Corp, publication
Porter, Arthur Kingsley The Construction of Lombard and Gothic
Vaults (New Haven, 1911) Yale University Press
Rohlfs, Gerhard
(Firenze, 1963)
Primitive Construzioni a Cupola in Europa
Sachse, H. J. Barocke Dachwerke, Decken und Gewlbe
(Berlin, 1975). Gebr. Mann Verlag Publishers
Sacriste, Eduardo;
Kechichian, Pedro Antonio;
Viviendas con Bovedas
(Buenos Aires, ?)
142
MacIntosh, Guillermo
Libreria Tecnica
Salvadori, Mario Why Buildings Stand Up "The Strength of
Architecture" (New York, 1982) McGraw-..Hill
Stechow, Wolfgang Rubens and the Classical Tradition
1968) Harvard University Press
(Cambridge,
Thompson, D'arcy W. On Growth and Form ed. John Tyler Bonner
(England, 1971) Hazell Watson & Viney Ltd.
Zalewski, Waclaw P.
Chapter 1
Chapter 2
Chapter 3
Chapter
Chapter
Chapter
Chapter
4
5
6
7
Notes on Structural Form:
Chapters 1 - 7
"Synthesis of Behavior of Structures"
"Stresses and Strains"
"Materials Used in Structures and Forces
Acting on Them"
"Linear Funicular Structures"
"Membranes and Shells"
"Simple Beams"
"Continuous Beams, Frames, Slabs"
Zalewski, Waclaw P. Formas Estructurales de Fuerza Constante
143
144
Illustrations
Final Design Competition Entry: Fall Term, 1982
Structural Engineering Laboratory 1,105J/4.315J
Author photographed
Fig. 1
Fan vaulting of King's College Chapel, completed 1515.
borrowed from: Leedy Jr. Walter C. Fan Vaulting Scientific
American
February, 1983
Vol. 248
No. 2 p.134
Fig. 2
Tamar at Saltash, Isambard Brunel, completed 1858.
borrowed from: Hayden, Martin The Book of Bridges (England, 1976)
Marshall Cavendish Publications p. 82
Fig. 3
Bayonne Bridge, New York State, completed 1932.
borrowed from: Hayden, Martin The Book of Bridges p. 95
Fig 1-7
Corrugated shell of the great sea clam.
Author photographed
Roman aqueduct in Segovia, Spain.
Fig. 2-6
borrowed from: Hayden, Martin The Book of Bridges
Fig. 2-7
p. 22
Snow bridge on fire escape in Somerville, winter 1983.
Author photographed
Fig. 4-2
Olmec sarcophagus from La Venta, Mexico, 800 - 400 B. C..
Author photographed
145
Fig. 4-3
Maya stone corbel arch, Palenque, Mexico, classic period
300 - 900 B. C..
Author photographed
Fig. 4-4
Modified Maya stone corbel arch, Chichen Itza Viejo,
Mexico,classic period 300 - 900 A. D..
Author photographed
Fig. 4-5a
Pension in Oaxaca, Mexico, built 1970.
Author photographed
Fig. 4-5b
Sewing factory, Paris, France, Auguste Perret, completed
1919.
borrowed from: Les Albums D'Art Druet XVI A. & G. Perret
Librairie de France, plate IX
(Paris, ?)
Fig. 4-10
Two hypars intersecting in ridges which diagonally
connect the four corners.
borrowed from: Gheorghin, Adrian; Dragomir, Virgil Geometry of
Structural Forms trans. Dinu Manoil (London, 1978) Applied Science
Publishers Ltd. Fig. 141 p. 135
Fig. 4-12
Conoid intersected by slanted plane to show negative
total curvature.
borrowed from: Zalewski, Waclaw P. Membranes and Shells Fig. 5.51
p. 37
Fig. 4-14
Ceiling/roof made from four hypars set against one
another.
borrowed from:
Zalewski, Waclaw P. Membranes and Shells
Fig. 4-17
Classic Gothic ribbed vault.
borrowed from: Leedy Jr., Walter C. Fan Vaulting
American February, 1983 Vol. 248 No. 2 p. 136
146
p. 35
Scientific
Fig 4-18
Fan vault created from fan-vault conoids.
borrowed from: Leedy Jr., Walter C. Fan Vaulting Scientific
American p. 136
Multi-story project for inustrial application
Fig. 4-19a and b
in Poland, 1958, Waclaw Zalewski, engineer.
borrowed from: Zalewski, Waclaw P. Fuerzas Estructurales de Fuerza
Constante
p. 78
Fig. 4-21a
Maria Steinbach, Germany, monumental wood truss
supporting roof.
borrowed from: Sachse, H. J. Barocke Dackwerke, Decken und Gewalbe
(Berlin, 1975) Gebr. Mann Verlag Publishers plate 186
Fig. C-1
"The Three Graces,"
1638 - 40, Rubens (Prado, Madrid).
borrowed from: Stechow, Wolfgang Rubens and the Classical Tradition
(Cambridge, 1968) Harvard University Press Fig. 39 p. 52
"The Adoration of the Shepherds," 1603 - 7, El Greco
Fig. C-2
(Prado, Madrid).
borrowed from: Gudiol, Jose El Greco 1541 - 1614 (New York, 1973)
The Viking Press Inc., Fig. 208 p. 232
147
148
149
150
151
152